Limits, Continuity and Differentiability

$ \mathrm{e}^{\mathrm{x}_0}+\mathrm{x}_0=0 $

$ \begin{aligned} & g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} \\ & g(x)=x-\frac{\alpha\left(e^x+x\right)}{3\left(e^x+1\right)}, g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \end{aligned} $

$ g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) $

$\Rightarrow$ As very clear $\mathrm{g}\left(\mathrm{x}_0\right)+\mathrm{x}_0 \Rightarrow 0$

$ \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| \frac{0 \uparrow}{0 \uparrow}=\lim \limits_{x \rightarrow x_0}\left|\frac{g^{\prime}(x)}{1}\right|=\left|g^{\prime}\left(x_0\right)\right| $

$ \mathrm{g}^{\prime}(\mathrm{x})=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+1\right)-\left(\mathrm{e}^{\mathrm{x}}+\mathrm{x}\right) \mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}+1\right)^2}\right) $

$ \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}\right)=1-\frac{\alpha}{3} $

$ \Rightarrow \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime}\left(x_0\right)\right|=\left|1-\frac{\alpha}{3}\right| $

The function $ f : \mathbb{R} \to \mathbb{R} $ is defined by:

$ f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x}, & \text{if } x \neq 0, \\ 2, & \text{if } x = 0. \end{cases} $

Explanation:

(A) To determine differentiability at $ x = 0 $, consider the right-hand derivative (RHD):

$ \lim\limits_{h \rightarrow 0} \frac{(2 - 2h^2 - h^2 \sin \frac{1}{h}) - 2}{h} = 0 $

Similarly, for the left-hand derivative (LHD) at $ x = 0 $, we also find it equals 0. Therefore, $ f $ is differentiable at $ x = 0 $.

(B) For $ x \in (0, \delta) $, the derivative of $ f $ is:

$ f^{\prime}(x) = -\left(4x + 2x \sin \frac{1}{x}\right) + \cos \frac{1}{x} $

Thus:

$ f^{\prime}(x) = -\left(2x\left(4 - \sin \frac{1}{x}\right)\right) + \cos \frac{1}{x} $

Since $\cos \frac{1}{x}$ oscillates, we cannot assert that $ f(x) $ is strictly decreasing on $ (0, \delta) $.

(C) For $ x \in (-\delta, 0) $ and any $ \delta > 0 $, $ f(x) $ is not increasing since $\cos \frac{1}{x}$ oscillates between -1 and 1.

(D) Evaluating at $ x = 0 $:

$ f(0) = 2 $

For small $ h $:

$ f(0 + h) < 2 \quad \text{and} \quad f(0 - h) < 2 $

Therefore, $ x = 0 $ is a local maximum, not a local minimum.

$\begin{aligned} & \text { (P) Let } g(x)=10 x^3-45 x^2+60 x+35 \\\\ & g^{\prime}(x)=30\left(x^2-3 x+2\right)=30(x-1)(x-2)\end{aligned}$

$\Rightarrow g(x)$ is decreasing in $[1,2]$

Now, $f(1)=\left[\frac{g(1)}{n}\right]=\left[\frac{60}{n}\right], f(2)=\left[\frac{55}{n}\right]$

For $f(x)$ to be continuous in $[1,2]$ its integral value should remain same in whole interval for $n=9, f(1)=6, f(2)=6$

$(P) \rightarrow(2)$

(Q)

$ \begin{aligned} & g^{\prime}(x)=\left(2 n^2-13 n-15\right)\left(3 x^2+3\right) \\ & =(2 n-15)(n+1)\left(3 x^2+3\right)>0 \end{aligned} $

for $g(x)$ to be increasing $\min n=8$

$\Rightarrow Q \rightarrow 1$ matching

(R) $h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right)$

$h(3)=0$ for $n>5$

at $n=6 \Rightarrow n(3+\delta)>h(3)$

$ n(3-\delta)>n(3) $

$\Rightarrow h(x)$ has local minima at $x=3$ for $n=6$

$R \rightarrow 4$ matching

(S) $\quad I(x)=\sin |x|+\cos \left|x+\frac{1}{2}\right|+\sin |x-1|+\cos \left|x-\frac{1}{2}\right|$ $+\ldots \ldots+\sin |x-4|+\cos \left|x-\frac{7}{2}\right|$

as $\sin |x-a|$ is non differentiable at $x=a$

but $\cos |x-a|$ remains differentiable at $x=a$

$\Rightarrow$ given $I(x)$ is non-differentiable at

$ \mathop {\lim }\limits_{x \to 0} \frac{(\cos x)^{\frac{1}{2}}-(\cos x)^{\frac{1}{3}}}{\sin ^2 x} $

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(\frac{1}{2 \sqrt{\cos x}} \times(-\sin x)\right)+\frac{1}{3 \times(\cos x)} \times \sin x}{2 \sin x \cdot \cos x} \\ & \lim _{x \rightarrow 0} \frac{\frac{-1}{2 \sqrt{\cos x}}+\frac{1}{3 \times(\cos x)^{\frac{2}{3}}}}{2 \cos x}=\frac{\frac{-1}{2}+\frac{1}{3}}{2}=\frac{-1}{12} \end{aligned} $

$f(x)=\left[x^2-3\right]$

Doubts points of $f(x)$ are at points where $x^2-3$ is integer

$ \begin{aligned} \Rightarrow \quad & x=0,1, \sqrt{2}, \sqrt{3} \\ & f\left(0^{+}\right)=\left[x^2\right]-3=-3 \\ & f\left(0^{-}\right)=\left[x^2\right]-3=-3 \\ & f(0)=-3 \end{aligned} $

Continuous at $x=0$

$ \begin{aligned} & f\left(0^{+}\right)=\left[x^2\right]-3=1-3=-2 \\ & f\left(0^{-}\right)=\left[x^2\right]-3=0-3=-3 \end{aligned} $

Discontinuous at $x=1$

$ \begin{aligned} & f\left(\sqrt{2}^{+}\right)=\left[x^2\right]-3=2-3=-1 \\ & f(\sqrt{2})=\left[x^2\right]-3=1-3=-2 \end{aligned} $

Discontinuous at $x=\sqrt{2}$

$ \begin{aligned} & f(\sqrt{3})^{+}=\left[x^2\right]-3=3-3=0 \\ & f(\sqrt{3})^{-}=\left[x^2\right]-3=2-3=-1 \end{aligned} $

Discontinuous at $x=\sqrt{3}$

Numbers of points of discontinuity $=3$

$f(x)=x^2\left|\cos \frac{\pi}{2}\right|, x \neq 0$

$ \begin{aligned} & =0, \quad x=0 \\ & f\left(2^{+}\right)=4\left|\cos \frac{\pi}{2}\right|=0 \\ & f\left(2^{-}\right)=4\left|\cos \frac{\pi}{2}\right|=0 \\ & f(2)=0 \end{aligned} $

$ \begin{aligned} & \text { Continuous at } x=2 \\ & f^{\prime}\left(2^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(2)}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{(h+2)^2\left|\cos \frac{\pi}{2+h}\right|}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{\left(h+2^2 \cos \left(\frac{\pi}{2+h}\right)\right.}{h} \\ & =\lim _{h \rightarrow 0^{+}} 2(h+2) \cos \left(\frac{\pi}{h+2}\right)+(h+2)^2 \\ & \quad x-\sin \left(\frac{\pi}{2+h}\right) \times \frac{-\pi}{(2+h)^2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =4 \cos \frac{\pi}{2}+\left(4 \times-\sin \frac{\pi}{2} \times \frac{-\pi}{4}\right)=\pi \\ & f^{\prime}(2) \lim _{h \rightarrow 0^{-}} \frac{f(2+h)-f(2)}{h} \\ & \quad=\lim _{h \rightarrow 0^{-}} \frac{(2+h)^2\left|\cos \frac{\pi}{2+h}\right|}{h} \\ & \quad=\lim _{h \rightarrow 0^{-}} \frac{-(2+h)^2 \cos \frac{\pi}{2+h}}{h} \\ & \quad-2(2+h) \cos \frac{\pi}{2+h}-(2+h)^2 \\ & \quad=\lim _{h \rightarrow 0^{-}} \frac{\times \sin \left(\frac{\pi}{2+h}\right) \times \frac{\pi}{(2+h)^2}}{1} \\ & \quad=0-\pi=-\pi \\ & f^{\prime}\left(2^{+}\right) \neq f^{\prime}(2) \end{aligned}\\ &\text { So, } f \text { is not differentiable at } x=2 \end{aligned} $

$|x|$ is non-differentiable at $x=0$

Also $||x|-1|$ will be non-differentiable at points where $|x|-1=0$

So, $||x|-1|$ is non-differentiable at $x=0, \pm 1$

So, set of all values of $x$ for which $||x|-1|$ is differentiable is $R-\{-1,1,0\}$

$ \begin{aligned} & \text { Let } \lim _{x \rightarrow 0} \frac{3^{x^3}-\left(1-x^3\right)^{2 / 3}}{x^2 \sin x}=L \\ & \Rightarrow \quad L=\lim _{x \rightarrow 0} \frac{3^{x^3}-\left(1-\frac{2}{3} x^3+\ldots\right)}{x^3\left(\frac{\sin x}{x}\right)} \\ & L=\lim _{x \rightarrow 0} \frac{3^{x^3}-1}{x^3 \cdot\left(\frac{\sin x}{x}\right)} +\frac{x^3\left(\frac{2}{3}+\text { terms containing } x\right)}{x^3 \times \frac{\sin x}{x}} \end{aligned} $

$ \begin{array}{rlr} L & =\ln 3+\frac{2}{3}=\ln q+p \\ q & =3, \\ p q & =2 \end{array} p=2 / 3 . $

$ \text { } \begin{aligned} & f(x)=2[x]-\frac{x}{|x|}, \quad x \neq 0 \\ &=1, x=0 \\ & f\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} 2[x]-\frac{x}{|x|} \\ &=\lim _{x \rightarrow 0^{+}} \frac{0-x}{x}=-1 \\ & f\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} 2[x]-\frac{x}{|x|} \\ & \lim _{x \rightarrow 0^{-}}-\left(-2-\frac{x}{-x}\right) \\ & f(0)=1\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, (given)\end{aligned} $

$ \begin{aligned} &\begin{aligned} f\left(1^{+}\right) & =\lim _{x \rightarrow 1^{+}} 2[x]-\frac{x}{|x|} \\ f\left(1^{+}\right) & =\lim _{x \rightarrow 1^{+}} 2-\frac{x}{x}=1 \\ f(1) & =2[1]-\frac{1}{|1|}=1 \end{aligned}\\ &\Rightarrow f \text { is right continuous at } x=1 \end{aligned} $

$ \begin{aligned} \text { } & \lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]} \\ \lim _{x \rightarrow 3} & =\frac{11-2-[-x]}{[x]+10} \\ \mathrm{LHL} & =\lim _{x \rightarrow 3^{-}} \frac{11-2-[-x]}{[x]+10}=\frac{9+3}{2+10}=1 \\ \mathrm{RHL} & =\lim _{x \rightarrow 3^{+}} \frac{9-[-x]}{[x]+10}=\frac{9+4}{3+10}=1 \\ \therefore & \text { limiting value }=1 \end{aligned} $

$ \text { } f(x)=\left[\begin{array}{cc} \frac{\cos 3 x-\cos x}{x \sin x}, & x<0 \\ p, & x=0 \\ \frac{\log (1+q \sin x)}{x}, & x>0 \end{array}\right. $

$ \begin{aligned} &\because f(x) \text { is continuous at } x=0\\ &\begin{aligned} & f\left(0^{-}\right)=f(0)=f\left(0^{+}\right) \\ & \lim _{x \rightarrow 0^{-}} \frac{\cos 3 x-\cos x}{x \sin x}=p \\ & \quad=\lim _{x \rightarrow 0^{+}} \frac{\log (1+q \sin x)}{x} \\ & \lim _{x \rightarrow 0^{-}} \frac{-2 \sin 2 x \times \sin (x)}{x \sin x}=p \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\log (1+q \sin x)(q \sin x)}{x(q \sin x)} \\ -4 & =p=q \\ \therefore p+q & =-8 \end{aligned} $

$ \begin{aligned} & \text { } \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}\left(1-\{x\}^3\right)} \\ & \text { If } x \rightarrow 0^{-},\{x\}=1+x \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-(1+x)^2\right) \sin ^{-1}(1-(1+x))}{(1+x)\left(1-(1+x)^3\right)} \\ & \quad=\lim _{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(-x^2-2 x\right) \sin ^{-1}(-x)}{(1+x)(-x)\left(x^2+3 x+3\right)} \\ & \quad=\frac{\cos ^{-1} 0}{1 \cdot 3}=\frac{\pi}{6} \\ & \therefore \quad \theta=\frac{\pi}{6} \Rightarrow \tan \theta=\frac{1}{\sqrt{3}} \end{aligned} $

$\because f(x)$ is continuous at $x=0$

$ \therefore f(0)=\lim\limits_{x \rightarrow 0} \frac{(a(625+x))^{1 / 5}-5}{(625+b x)^{1 / 4}-5} $

at $x=0$

$\frac{(625 a)^{1 / 5}-5}{0}$ (limit exist only when

$ \begin{aligned} & (625 a)^{1 / 5}-5=0 \\ & \therefore \quad a=5) \end{aligned} $

Using L'Hospital rule,

$ \begin{aligned} f(0)= & \lim _{x \rightarrow 0} \frac{\frac{1}{5}\left(a(625+x)^{-4 / 5}\right) \cdot a}{\frac{1}{4}(625+b x)^{-3 / 4} \cdot b} \\ & =\frac{\frac{1}{5}(a \cdot 625)^{-4 / 5} a}{\frac{1}{4}(625)^{-3 / 4} b}=\frac{4}{5} \frac{\left(5^5\right)^{-4 / 5}}{\left(5^4\right)^{-3 / 4}} \cdot \frac{5}{b} \\ & =\frac{4 \cdot 5^{-4}}{5^{-3} \cdot b}=\frac{4}{5 b} \end{aligned} $

$ \begin{aligned} & \text { } f(x)=7|2 x+1|-19|3 x-5| \\ & f(x)= \begin{cases}-7(2 x+1)+19(3 x-5), & x<-\frac{1}{2} \\ 7(2 x+1)+19(3 x-5), & -\frac{1}{2} \leq x<\frac{5}{3} \\ 7(2 x+1)-19(3 x-5), & x \geq \frac{5}{3}\end{cases} \end{aligned} $

$ \begin{aligned} & f(x)=\left\{\begin{array}{l} 43 x-102, x \leq \frac{-1}{2} \\ 71 x-88,-\frac{1}{2} \leq x < \frac{5}{3} \\ -43 x+102, x \geq \frac{5}{3} \end{array}\right. \\ & f^{\prime}(x)=43\left\{\begin{array}{l} 43, x < \frac{-1}{2} \\ 71, \frac{-1}{2} < x < \frac{5}{3} \\ -43, x>\frac{5}{3} \end{array}\right. \end{aligned} $

$ \therefore f(x) \text { is differentiable } x=\frac{-1}{2}, \frac{5}{3} $

$ \begin{aligned} &\text { Given, } f(x)=\frac{x\left(a^x-1\right)}{1-\cos x}\\ &\begin{aligned} & \text { And } g(x)=\frac{x\left(1-a^x\right)}{a^x\left(\sqrt{1-x^2}-\sqrt{1+x^2}\right)} \\ & \begin{aligned} \therefore \lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0} \frac{x\left(a^x-1\right)}{1-\cos x} \\ & =\lim _{x \rightarrow 0} \frac{x \frac{\left(a^x-1\right)}{x}}{\frac{1-\cos x}{x}} \\ & =\lim _{x \rightarrow 0} \frac{a^x-1}{x} \times \frac{x^2}{1-\cos x} \\ & =\lim _{x \rightarrow 0} \frac{a^x-1}{x} \times \lim _{x \rightarrow 0} \frac{x^2}{1+\cos x} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\log a \times 2 \\ & =2 \log a \end{aligned}\\ &\text { Also, } \lim _{x \rightarrow 0} g(x)\\ &\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x\left(1-a^x\right)}{a^x\left(\sqrt{1-x^2}-\sqrt{1+x^2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{x\left(1-a^x\right)\left(\sqrt{1-x^2}+\sqrt{1+x^2}\right)}{a^x\left(\sqrt{1-x^2}-\sqrt{1+x^2}\right)} \\ & \left(\sqrt{1-x^2}+\sqrt{1+x^2}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{x\left(1-a^x\right)\left(\sqrt{1-x^2}+\sqrt{1-x^2}\right)}{a^x\left(1-x^2-1+x^2\right)} \\ & =\lim _{x \rightarrow 0} \frac{x\left(1-a^x\right)\left(\sqrt{1-x^2}+\sqrt{1-x^2}\right)}{a^x\left(-2 x^2\right)} \\ & =\lim _{x \rightarrow 0} \frac{\left(a^x-1\right)\left(\sqrt{1-x^2}+\sqrt{1-x^2}\right)}{a^x(2 x)} \\ & =\lim _{x \rightarrow 0} \frac{a^x-1}{x} \times \lim _{x \rightarrow 0} \frac{\sqrt{1-x^2}+\sqrt{1-x^2}}{2 a^x} \\ & =\log a \times \frac{1+1}{2(1)} \\ & =\log _a a(1)=\log a \\ & \therefore \lim _{x \rightarrow 0}(f(x)-g(x)) \\ & =\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0} g(x) \\ & =2 \log a-\log a=\log a \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{c} \frac{a \sin x-b x+c x^2+x^3}{2 \log (1+x)-2 x^3+x^4}, x \neq 0 \\ 0, \quad x=0 \end{array}\right. $

is continuous at $x=0$

$ \begin{aligned} & \therefore \lim _{x \rightarrow 0} f(x)=f(0) . \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{a \sin x-b x+c x^2+x^3}{2 \log (1+x)-2 x^3+x^4}\left(\frac{0}{0}\right)=f(0) \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{a \cos x-b+2 c x+3 x^2}{\frac{2}{(1+x)}-6 x^2+4 x^3}=0 \\ & \Rightarrow \quad \frac{a-b}{2}=0 \Rightarrow a-b=0 \\ & \Rightarrow \quad a=b \end{aligned} $

Given, $g(x)=\left\{\begin{array}{ll}K \sqrt{x+1}, & 0 \leq x \leq 3 \\ m x+2, & 3 < x \leq 5\end{array}\right.$ is differentiable

Since, $g(x)$ is differentiable therefore it is continuous.

at $x=3$

$ \begin{aligned} & \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x) \\ \Rightarrow & \lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} f(3+h) \\ \Rightarrow & \lim _{h \rightarrow 0}(K \sqrt{(3-h+1)})=\lim _{h \rightarrow 0}(m(3+h)+2) \\ \Rightarrow & K \sqrt{4}=m(3)+2 \end{aligned} $

$\Rightarrow \quad 2 K=3 m+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

for $0 \leq x \leq 3$

$ g^{\prime}(x)=K \frac{1}{2 \sqrt{x+1}} $

And for $3

$ g^{\prime}(x)=m $

for differentiability at $x=3$

$ \begin{aligned} & L g^{\prime}(3)=R g^{\prime}(3) \\ & \Rightarrow \frac{K}{2 \sqrt{4}}=m \\ \Rightarrow & K=4 m \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On Solving Eqs. (i) and (ii), we get

$ \begin{aligned} & & 2(4 m) & =3 m+2 \\ \Rightarrow & & 8 m & =3 m+2 \\ \Rightarrow & & 5 m & =2 \Rightarrow m=\frac{2}{5} \end{aligned} $

And $\quad K=\frac{8}{5}$

$ \therefore \quad K+m=\frac{8}{5}+\frac{2}{5}=\frac{10}{5}=2 $

Since, $x \rightarrow 3^{-}$, we have $x<3$.

$ \therefore|x|=x \text { and }|3-x|=3-x $

As, $x \rightarrow 3^{-}, 3 x \rightarrow 9^{-}$, so $9-3 x \rightarrow 0^{+}$

Thus, $[9-3 x]=0$ for $x$ slightly less than 3 and $[3 x-9]=-1$ for $x$ slightly less than 3 .

$ \begin{aligned} \therefore \lim _{x \rightarrow 3^{-}} & \frac{(3-x+\sin (3-x)) \cos (0)}{(3-x)(-1)} \\ & =\lim _{x \rightarrow 3^{-}} \frac{(3-x+\sin (3-x))}{(3-x)(-1)} \end{aligned} $

Let $t=3-x$.

As, $x \rightarrow 3^{-}, t \rightarrow 0^{+}$

$ \begin{aligned} & \text { So, } \quad \lim _{x \rightarrow 3^{-}} \frac{(3-x+\sin (3-x))}{(3-x)(-1)} \\ & \Rightarrow \quad \lim _{t \rightarrow 0^{+}} \frac{t+\sin t}{-t} \\ & \Rightarrow \quad \lim _{t \rightarrow 0^{+}}\left(-1-\frac{\sin t}{t}\right) \\ & \Rightarrow \quad-1-1 \quad\left[\because \lim _{t \rightarrow 0} \frac{\sin t}{t}=1\right] \\ & \Rightarrow \quad-2 \\ & \therefore \lim _{x \rightarrow 3^{-}} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]} \\ & \quad=-2 \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{cc} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)}, & x \neq 0 \\ \log 3 \log 4, & x=0 \end{array}\right. $

Now, $f(0)=\log 3 \log 4$

Now, $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)}$

$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{\left(3^x-1\right)\left(2^x-1\right)}{1-\cos \left(\frac{x}{a}\right)} $

$ \begin{aligned} & \Rightarrow \lim _{x \rightarrow 0}\left[\frac{\frac{3^x-1}{x} \cdot \frac{2^x-1}{x}}{\frac{1-\cos \left(\frac{x}{a}\right)}{x^2}}\right] \\ & \Rightarrow \frac{(\ln 3)(\ln 2)}{\frac{1}{\left(2 a^2\right)}}\left[\begin{array}{l} \because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\ln a \\ \text { and } \lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{k^2}{2} \end{array}\right] \\ & =2 a^2(\ln 3) \ln (2) \end{aligned} $

Since, given function is continuous

So, $\mathop {\lim }\limits_{x \to 0} f(x)=f(0)$

$ \begin{aligned} & \Rightarrow \quad 2 a^2(\ln 3)(\ln 2)=(\ln 3)(\ln 4) \\ & \Rightarrow \quad(\ln 3)(\ln 2)^2=2(\ln 3)(\ln 2) \\ & \Rightarrow \quad 2 a^2=2 \Rightarrow a^2=1 \end{aligned} $

$\Rightarrow \quad a=1 \quad(\because a$ is a positive real number $)$

$ \begin{aligned} & \text { } \lim _{x \rightarrow \infty}[x-\log (\cosh x)] \\ & \because \quad \cosh x=\frac{e^x+e^{-x}}{2} \end{aligned} $

$\because x \rightarrow \infty, e^{-x}$ become negligible

$ \begin{aligned} \therefore \cosh x & \simeq \frac{e^x}{2} \\ \Rightarrow \log (\cosh x) & =\log \left(\frac{e^x}{2}\right)=x-\log 2 \\ & =\lim _{x \rightarrow \infty}[x-x+\log 2]=\log 2 \end{aligned} $

$ \begin{aligned} & \left.\text { } \lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)\right. \\ & =\lim _{x \rightarrow \infty}\left(x\left(\sqrt[3]{1+\frac{4}{x}}-\sqrt{1-\frac{3}{x}}\right)\right) \end{aligned} $

$ \begin{aligned} &\text { Apply binomial expansion for radicals }\\ &\begin{aligned} & =\lim _{x \rightarrow \infty}\left(x \left(\left(1+\frac{1}{3} \cdot \frac{4}{x}+0\left(\frac{1}{x^2}\right)\right)\right.\right. \left.-\left(1-\frac{1}{2} \cdot \frac{3}{x}+0\left(\frac{1}{x^2}\right)\right)\right) \\ & =\lim _{x \rightarrow \infty} x\left(1+\frac{4}{3 x}-1+\frac{3}{2 x}\right) \\ & =\lim _{x \rightarrow \infty} x \cdot \frac{1}{x}\left(\frac{4}{3}+\frac{3}{2}\right) \\ & =\frac{4}{3}+\frac{3}{2}=\frac{8+9}{6}=17 / 6 \end{aligned} \end{aligned} $

$\because f(x)$ is continuous at $x=1$

$ \therefore \mathop {\lim }\limits_{x \to {1^ - }} f(x)=f(1)=\mathop {\lim }\limits_{x \to {1^ +}} f(x) $

So, $\mathop {\lim }\limits_{x \to {1^ - }} f(x)=\mathop {\lim }\limits_{x \to {1^ - }} \left[e^{\frac{\sin a(x-[x])}{x-[x]}}\right]$

$ =\mathop {\lim }\limits_{x \to {1^ - }} \left(e^{\frac{\sin a(x-0)}{x-0}}\right)=e^{\sin a} $

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x)=\mathop {\lim }\limits_{x \to {1^+ }} \frac{\left|x^2+x-2\right|}{x-1}$

$ =\mathop {\lim }\limits_{x \to {1^+ }} \frac{(x-1)(x+2)}{(x-1)}=\mathop {\lim }\limits_{x \to {1^+ }}(x+2)=3 $

and $\quad f(1)=b+1=3$

$\Rightarrow b=2$ and $e^{\sin a}=3 \Rightarrow \sin a=\ln 3$

Hence, $b \sin a=2 \ln 3=\ln 9$

We have,

$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}\left(\frac{0}{0} \text { form }\right) $

Using L Hospital rule,

$ \begin{aligned} & \lim _{x \rightarrow \frac{\pi}{4}} \frac{-3(\cos x+\sin x)^2(\cos x-\sin x)}{-2 \cos 2 x} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-3(\cos x+\sin x)\left(\cos ^2 x-\sin ^2 x\right)}{-2(\cos 2 x)} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{3(\cos x+\sin x)(\cos 2 x)}{2 \cos 2 x} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{3}{2}(\cos x+\sin x) \\ & =\frac{3}{2} \times \sqrt{2}=\frac{3}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } \lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right) \\ & \lim _{x \rightarrow 2^{+}}\left(\frac{\left[2^{+}\right]^3}{3}-\left[\frac{2^{+}}{3}\right]^3\right)=\frac{8}{3}-0=\frac{8}{3} \end{aligned} \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{cl} \frac{1-\cos 4 x}{x^2}, & x<0 \\ a & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{array}\right. $

$\because f(x)$ is continuous at $x=0$

$ \begin{aligned} \therefore f(0) & =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ a & =\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \\ \Rightarrow a & =\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 2 x}{x^2} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4}{16+\sqrt{x}-16} \\ \Rightarrow \quad a & =\lim _{x \rightarrow 0^{-}} 8\left(\frac{\sin 2 x}{2 x}\right)^2 \\ & =\lim _{x \rightarrow 0^{+}} \sqrt{16+\sqrt{x}}+4 \\ \quad a & =8=8 \\ \Rightarrow \quad a & =8 \end{aligned} $

$ \text { } \begin{aligned} f(x) & =\frac{x}{1+|x|} \\ & =\left[\begin{array}{ll} \frac{x}{1-x}, & x<0 \\ \frac{x}{1+x}, & x \geq 0 \end{array}\right. \end{aligned} $

∵ Given function is differentiable for all real $x$ but we have to check at $x=0$

$ \begin{aligned} & \begin{aligned} \text { LHD } & =f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{\frac{-h}{1+h}-0}{-h}=1 \end{aligned} \\ & \text { RHD } =f^{\prime}\left(0^{+}\right) =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & \Rightarrow \lim _{h \rightarrow 0} \frac{\frac{h}{1+h}-0}{h}=1 \end{aligned} $

$ \because L H D=R H D $

∴ Differentiable at $x=0$

$\therefore f(x)$ is differentiate for $x \in(-\infty, \infty)$

$ \mathop {\lim }\limits_{x \to 0} \frac{x+2 \sin x+3 \tan x-\tan ^3 x}{\sqrt{x^2+2 \sin x+\tan x+3-\sqrt{\sin ^2 x-2 \tan x-x+3}}} $

$ \begin{aligned} &\text { After rationalisation }\\ &\left(x+2 \sin x+3 \tan x-\tan ^3 x\right) \end{aligned} $

$ \mathop {\lim }\limits_{x \to 0} \frac{\left(\sqrt{x^2+2 \sin x+\tan x+3}\right.+\sqrt{\sin ^2 x-2 \tan x-x+3}}{x^2+2 \sin x+\tan x+3-\sin ^2 x+2 \tan x+x-3} $

$ x\left(1+\frac{2 \sin x}{x}+\frac{3 \tan x}{x}-\frac{3 \tan ^3 x}{x}\right) $

$ \mathop {\lim }\limits_{x \to 0} \frac{\left.\left(\sqrt{x^2+2 \sin x+\tan x+3}\right.+\sqrt{\sin ^2 x-2 \tan x-x+3}\right)}{x\left(1+x+\frac{2 \sin x}{x}+\frac{\tan x}{x}\right.\left.-\frac{\sin ^2 x}{x}+\frac{2 \tan x}{x}\right)} $

$ \Rightarrow \frac{(1+2+3)(\sqrt{3}+\sqrt{3})}{(1+2+1+2)}=\frac{6 \times 2 \sqrt{3}}{6}=2 \sqrt{3} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{x \rightarrow \infty} \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}} \\ & =\lim _{x \rightarrow \infty} \frac{x^{25} \cdot x^{35}\left(\frac{3}{x}-1\right)^{25}\left(\frac{6}{x}+1\right)^{35}}{x^{38} \cdot x^{22}\left(\frac{12}{x}+1\right)^{38}\left(\frac{9}{x}-1\right)^{22}} \\ & =\lim _{x \rightarrow \infty} \frac{\left(\frac{3}{x}-1\right)^{25}\left(\frac{6}{x}+1\right)^{35}}{\left(\frac{12}{x}+1\right)^{38}\left(\frac{9}{x}-1\right)^{22}} \end{aligned} \end{aligned} $

$ =\frac{(-1)(1)}{(1)(1)}=-1\left\{\begin{array}{c} x \rightarrow \infty \\ \frac{3}{x} \rightarrow 0 \\\because \frac{6}{x} \rightarrow 0 \\ \frac{12}{x} \rightarrow 0 \\ \frac{9}{x} \rightarrow 0 \end{array}\right\} $

We have,

$ f(x)=\left\{\begin{array}{cc} \log (1+[x]) & x \geq 0 \\ \sin ^{-1}[x] & -1 \leq x<0 \\ k([x]+|x|] & x<-1 \end{array}\right. $

$\because f(x)$ is continuous at $x=-1$

$ \begin{aligned} & \therefore f(-1)=f\left(-1^{-}\right)=f\left(-1^{+}\right) \\ & \sin ^{-1}(-1)=k\left(\left[-1^{-}\right]+\left|-1^{-}\right|\right)=\sin ^{-1}\left[-1^{+}\right] \\ & \Rightarrow \quad \frac{-\pi}{2}=k(-2+1)=\frac{-\pi}{2} \\ & \Rightarrow \quad-k=-\pi / 2 \\ & \therefore \quad k=\frac{\pi}{2} \end{aligned} $

$ \text { } \begin{aligned} \lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}\right. & +\sin \frac{3 \pi}{2 n} \left.+\ldots+\sin \frac{\pi}{2}\right] \\ = & \lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left(\frac{\sin \frac{n}{2}\left(\frac{\pi}{2 n}\right)}{\sin \frac{1}{2}\left(\frac{\pi}{2 n}\right)} \cdot \sin \left(\frac{\frac{\pi}{2 n}+\frac{\pi}{2}}{2}\right)\right) \\ = & \lim _{n \rightarrow \infty} \frac{\pi}{2 n} \cdot \frac{\left(\sin \frac{\pi}{4}\right)}{\sin \frac{\pi}{4 n}} \cdot \sin \left(\frac{\pi}{4 n}+\frac{\pi}{4}\right) \end{aligned} $

$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{\pi}{2 n} \frac{(1 / \sqrt{2})}{\left(\frac{\sin \frac{\pi}{4 n}}{\frac{\pi}{4 n}}\right)} \cdot \sin \left(\frac{\pi}{4 n}+\frac{\pi}{4}\right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& =2 \cdot \frac{1}{\sqrt{2}} \sin \pi / 4=1 \end{aligned} $

Since, $x \rightarrow 0^{+}$, so, the greatest integer function $[x]$ equals 0 as $0 \leq x<1$.

$ \Rightarrow \quad \lim\limits_{x \rightarrow 0^{+}} \frac{\cos (0)-\cos (k x-0)}{x^2}=5 \Rightarrow $

$ \begin{aligned} & \lim\limits _{x \rightarrow 0^{+}} \frac{1-\cos (k x)}{x^2}=5 \\ & \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{2 \sin ^2\left(\frac{k x}{2}\right)}{x^2}=5 \\ & \Rightarrow \lim _{x \rightarrow 0^{+}} 2 \frac{\sin ^2\left(\frac{k x}{2}\right)}{\left(\frac{k x}{2}\right)^2} \cdot \frac{k^2}{4}=5 \\ & \Rightarrow \quad 2 \cdot 1 \cdot \frac{k^2}{4}=5 \quad\left[\because \lim _{\theta \rightarrow 0} \frac{\sin (\theta)}{\theta}=1\right] \end{aligned} $

$ \begin{aligned} &\Rightarrow \quad k^2=\frac{20}{2} \Rightarrow k= \pm \sqrt{10}\\ &\text { So, } k=\sqrt{10} \end{aligned} $

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2} \\ & \quad=\lim _{x \rightarrow 0} \frac{x \cdot \frac{2 \tan x}{1-\tan ^2 x}-2 x \tan x}{\left(1-\left(1-2 \sin ^2 x\right)\right)^2} \\ & \quad=\lim _{x \rightarrow 0} \frac{2 x \tan x\left(\frac{1}{1-\tan ^2 x}-1\right)}{\left(2 \sin ^2 x\right)^2} \\ & \quad=\lim _{x \rightarrow 0} \frac{\frac{2 x \tan x\left(1-1+\tan ^2 x\right)}{1-\tan ^2 x}}{4 \sin ^4 x}=\lim _{x \rightarrow 0} \frac{2 x \tan x \cdot \tan ^2 x}{4 \sin ^4 x\left(1-\tan ^2 x\right)} \\ & \quad=\lim _{x \rightarrow 0} \frac{x \tan ^3 x}{2 \sin ^4 x\left(1-\tan ^2 x\right)}=\lim _{x \rightarrow 0} \frac{x \frac{\sin ^3 x}{\cos ^3 x}}{2\left(1-\tan ^2 x\right) \cdot \cos ^2 x \cdot \sin ^2 x} \\ & \quad=\lim _{x \rightarrow 0} \frac{1}{2 \cos ^3 x\left(1-\tan ^2 x\right)} \cdot \lim _{x \rightarrow 0} \frac{x}{\sin x} \end{aligned} $

$ \begin{array}{ll} =\frac{1}{2 \cos ^3 0\left(1-\tan ^2 0\right)} \cdot 1 & {\left[\because \lim _{x \rightarrow 0} \frac{x}{\sin x}=1\right]} \\ =\frac{1}{2(1)(1)}=\frac{1}{2} & \end{array} $

Given, function

$ f(x)=\left\{\begin{array}{cc} \frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x} & , \quad x>0 \\ \frac{\cos 4 x-\cos b x}{\tan ^2 x} & , \quad x=0 \end{array}\right. $

is contiuous at $x=0$

Since, $f(x)$ is continuous at $x=0$

So, $\mathop {\lim }\limits_{x \to 0 - } f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$ and $f(0)=2$

Now, $\mathop {\lim }\limits_{x \to 0 + } \frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x}$

Applying L'hospital rule, twice, since it is an indeterminate form $\frac{0}{0}$.

So, $\mathop {\lim }\limits_{x \to 0 + } \frac{a e^{a x} \log (1+x)+\frac{\left(e^{a x}-1\right)}{1+x}}{2 \cos x \sin x}$

This limit is again $\frac{0}{0}$ form, so apply L'Hospital rule again.

$\mathop {\lim }\limits_{x \to 0 + } \frac{a^2 e^{a x} \log (1+x)+\frac{a e^{a x}(1+x)-\left(e^{a x}-1\right)}{(1+x)^2}+\frac{a e^{a x}}{1+x}}{2 \cos ^2 x-2 \sin ^2 x}$

$ =\frac{0+\frac{a(1)-(1-1)}{1^2}+\frac{a}{1}}{2(1)-0}=\frac{a+a}{2}=a $

Since, $\mathop {\lim }\limits_{x \to 0 + } f(x)=f(0) \Rightarrow a=2$

Now, $\mathop {\lim }\limits_{x \to 0 - } \frac{\cos 4 x-\cos b x}{\tan ^2 x}$

$ =\mathop {\lim }\limits_{x \to 0 - } \frac{-2 \sin \left(\frac{4 x+b x}{2}\right) \sin \left(\frac{4 x-b x}{2}\right)}{\tan ^2 x} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{-2 \sin \left(\frac{4 x+b x}{2}\right) \sin \left(\frac{4 x-b x}{2}\right) \cdot \cos ^2 x}{\frac{\sin ^2 x}{x^2} \cdot x^2} \\ & \lim _{x \rightarrow 0^{-}} \frac{\frac{-2 \sin \left(\frac{4 x+b x}{2}\right)}{\frac{4 x+b x}{2}} \cdot\left(\frac{4 x+b x}{2}\right) \cdot \frac{\sin \left(\frac{4 x-b x}{2}\right)}{\left(\frac{4 x-b x}{2}\right)}\left(\frac{4 x-b x}{2}\right) \cdot \cos ^2 x}{\frac{\sin ^2 x}{x^2} \cdot x^2} \\ & =\lim _{x \rightarrow 0^{-}} \frac{-2 \cdot 1 \cdot\left(\frac{4 x+b x}{2}\right) \cdot 1 \cdot\left(\frac{4 x-b x}{2}\right) \cdot \cos ^2 x}{1 \cdot x^2} \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ & =\lim _{x \rightarrow 0^{-}} \frac{-2\left(16 x^2-b^2 x^2\right)}{4 x^2} \cdot \cos ^2 x \\ & =\lim _{x \rightarrow 0^{-}} \frac{\left(b^2-16\right)}{2} \cdot \cos ^2 x=\frac{b^2-16}{2} \cdot 1 \quad\left[\because \lim _{x \rightarrow 0^{-}} \cos ^2 x=1\right] \end{aligned} $

$ =\frac{b^2-16}{2} $

Also, $\mathop {\lim }\limits_{x \to 0 - } f(x)=f(0)$

$ \Rightarrow \quad \frac{b^2-16}{2}=2 \Rightarrow b^2=4+16=20 $

Now, $\sqrt{b^2-a^2}=\sqrt{20-2^2}=\sqrt{20-4}=\sqrt{16}=4$

$ \text { } \begin{aligned} & \lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2} \\ & =\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{\left(2 \sin ^2\left(\frac{3 x}{2}\right)\right)^2} \end{aligned} $

For small-angles $\{\because x \rightarrow 0\}$

$ \begin{aligned} & \sin k x \rightarrow k x \\ & =\lim _{x \rightarrow 0} \frac{x^2(3 x)^2+(6 x)^4}{\left(2\left(\frac{3 x}{2}\right)^2\right)^2} \\ & =\lim _{x \rightarrow 0} \frac{x^4[9+1296]}{x^4\left[\frac{9}{2}\right]^2}=\lim _{x \rightarrow 0} \frac{1305}{81} \times 4 \\ & =\frac{435}{27} \times 4=\frac{580}{9} \end{aligned} $

$ f(x)=\left\{\begin{array}{cc} (1+\sin x)^{\cos x}, & -\frac{\pi}{2} < x < 0 \\ a, & x=0 \\ \frac{e^{\frac{2}{x}}+e^{\frac{3}{x}}}{a e^{\frac{2}{x}}+b e^{\frac{3}{x}}}, & 0 < x < \frac{\pi}{2} \end{array}\right. $

So, $\mathop {\lim }\limits_{x \to {0^ - }} f(x)=\lim _{x \rightarrow 0^{-}}(1+\sin x)^{\cos x}$

Let $y=(1+\sin x)^{\cos x}$

$ \begin{aligned} \Rightarrow \ln y & =\cos x \ln (1+\sin x) \\ & =\frac{\ln (1+\sin x)}{\sin x}=1 \end{aligned} $

$\Rightarrow \quad y=e$

$ \therefore \mathop {\lim }\limits_{x \to {0^ - }}(1+\sin x)^{\cos x}=e=a=f(0) $

and $\mathop {\lim }\limits_{x \to {0^ +}} f(x)=\mathop {\lim }\limits_{x \to {0^+ }} \frac{e^{\frac{2}{x}}+e^{\frac{3}{x}}}{a e^{\frac{2}{x}}+b e^{\frac{3}{2}}}$

$ =\mathop {\lim }\limits_{x \to {0^ + }} \frac{e^{\frac{3}{x}}\left(e^{-\frac{1}{x}}+1\right)}{e^{\frac{3}{x}}\left(e \cdot e^{\frac{-1}{x}}+b\right)}=\frac{1}{b} $

So, $e=\frac{1}{b} \Rightarrow b=\frac{1}{e}$

$\mathop {\lim }\limits_{x \to 0} \frac{(\operatorname{cosec} x-\cot x)\left(e^x-e^{-x}\right)}{\sqrt{3}-\sqrt{2+\cos x}}$

Convert, $\operatorname{cosec} x$ and $\cot x$ into $\sin x$ and $e^x-e^{-x}=2 \sin h x$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\left(\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}} \cdot 2 \sin h x\right)}{\sqrt{3}-\sqrt{2+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{2 \tan \frac{x}{2} \cdot \sin h x}{\sqrt{3}-\sqrt{2+\cos x}} \times \frac{\sqrt{3}+\sqrt{2+\cos x}}{\sqrt{3}+\sqrt{2+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{2 \tan \frac{x}{2} \cdot \sin h x(\sqrt{3}+\sqrt{2+\cos x})}{1-\cos x} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{2 \tan \frac{x}{2} \cdot \sin h x(\sqrt{3}+\sqrt{2+\cos x})}{1-\cos x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{2 \sin \frac{x}{2}}{\cos \frac{x}{2}} \cdot \sin h x(\sqrt{3}+\sqrt{2+\cos x})}{2 \sin ^2 \frac{x}{2}} \\ & =\lim _{x \rightarrow 0} \frac{\sin h x(\sqrt{3}+\sqrt{2+\cos x}}{\cos \frac{x}{2} \cdot \sin \frac{x}{2}} \\ & \because x \rightarrow 0 \Rightarrow \cos \frac{x}{2} \rightarrow 1 \text { and } \sin \frac{x}{2} \rightarrow \frac{x}{2} \\ & \sin h x \rightarrow x \\ & \quad=\frac{x(\sqrt{3}+\sqrt{2+1})}{\frac{x}{2}}=2(\sqrt{3}+\sqrt{3}) \\ & \quad=2(2 \sqrt{3})=4 \sqrt{3} \end{aligned} $

$l=\mathop {\lim }\limits_{\theta \to 0} 3\left(\frac{\sin \theta}{\theta}\right)-\mathop {\lim }\limits_{\theta \to 0} 4 \cdot\left(\frac{\sin \theta}{\theta}\right)^3 \cdot \theta^2=3 \times 1-4 \times 0=3$

$ m=\mathop {\lim }\limits_{\theta \to 0}\left(\frac{2 \tan \theta}{\theta} \times \frac{1}{1-\tan ^2 \theta}\right)=2 \times 1 \times\left(\frac{1}{1-0}\right)=2 $

$\therefore$ The quadratic equation whose roots are 3 and 2 is

$ x^2-5 x+6=0 . $

$ \begin{aligned} & \text { Let } L=\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}} \\ & \left.=\lim _{x \rightarrow \infty} \frac{x\left(3+\frac{4 \cos ^2 x}{x}\right)}{x \sqrt{1-5\left(\frac{\sin x}{x}\right)^2}}\left(\because \lim _{x \rightarrow \infty}\left(\frac{\sin x}{x}\right)^2=\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{1}{y}\right)}{\frac{1}{y}}\right)\right)^2\right) \\ & =\lim _{y \rightarrow 0}\left(y \sin \left(\frac{1}{y}\right)\right)^2=(0 \times a \text { finite quantity })=0=\frac{3+0}{\sqrt{1-0}}=3 \end{aligned} $

LHL

$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{\left(1+a x^2+b x^3\right)^{\frac{1}{3}}-\left(1-a x^2-b x^3\right)^{\frac{1}{3}}}{x^2}\left(\text { form } \frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0^{-}} \frac{\frac{1}{3}\left(1+a x+b x^3\right)^{\frac{1}{3}-1}\left(2 a x+3 b x^2\right)-\frac{1}{3}\left(1-a x^2-b x^3\right)^{\frac{1}{3}-1}(-2 a x-3 b x)}{2 x} \end{aligned} $

(Using L Hospital rule)

$ =\frac{\frac{1}{3} \times(1+0+0)^{\frac{-2}{3}}(2 a+0)-\frac{1}{3}(1-0-0)^{\frac{-2}{3}}(-2 a-0)}{2} $

(cancel $x$ from both $N^r \times \operatorname{Dr}$ )

$ =\frac{\frac{2 a}{3}+\frac{2 a}{3}}{2}=\frac{4 a}{6}=\frac{2 a}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} &\text { RHL }\\ &\left(3 x+\frac{(3 x)^3}{3}+\frac{2(3 x)^5}{15}+\ldots\right) \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\tan 3 x-\sin 3 x}{b x^3}=\lim _{x \rightarrow 0^{+}} \frac{-\left(3 x-\frac{(3 x)^3}{3!}+\ldots\right)}{b x^3} \\ & \quad=\lim _{x \rightarrow 0^{+}} \frac{\frac{27 x^3}{3}+\frac{27 x^3}{6}+\text { (higher terms are zero) }}{b x^3}=\frac{27}{2 b} \\ & f(0)=5 \\ & \because \quad f(x) \text { is continuous at } x=0 . \\ & \therefore \quad \frac{2 a}{3}=5=\frac{27}{2 b} \end{aligned} $

$ \begin{aligned} &\Rightarrow \quad a=\frac{15}{2}, b=\frac{27}{10}\\ &\therefore \text { Geometric mean of } a \text { and } b=\sqrt{a b}=\sqrt{\frac{15}{2} \times \frac{27}{10}}=\frac{9}{2} \end{aligned} $

Evaluate $[x]$ as $x \rightarrow 0^{-}$

As $x$ approaches 0 from the left $[x]=-1 \{x\}$ as $x \rightarrow 0^{-}$

$ \{x\}=x-[x]=x-(-1)=x+1 $

as $x \rightarrow 0^{-},\{x\} \rightarrow 1$

$ \begin{aligned} \lim _{x \rightarrow 0^{-}} & \frac{\sin ^{-1}(x+[x])}{2-\{x\}} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x-1)}{2-(x+1)} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x-1)}{1-x} \end{aligned} $

as $x \rightarrow 0^{-}$, the limit becomes

$ \frac{\sin ^{-1}(-1)}{1-0}=-\frac{\pi}{2}=-\frac{\pi}{2} $

Therefore, $\theta=-\frac{\pi}{2}$

$ \begin{aligned} \sin \theta+\cos \theta & =\sin \left(-\frac{\pi}{2}\right)+\cos \left(-\frac{\pi}{2}\right) \\ & =-1 \\ \sin \theta+\cos \theta & =-1 \end{aligned} $

$ \text { } \begin{aligned} & \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x \\ = & \lim _{n \rightarrow \infty} \frac{1}{n^3} \frac{n(n+1)(2 n+1) x}{6} \\ = & \lim _{n \rightarrow \infty} \frac{n \cdot n \cdot n\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right) x}{6 n^3} \\ = & \frac{x}{6} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right) \\ = & \frac{x}{6} \times 2=\frac{x}{3} \end{aligned} $

Since, $f(x)$ is continuous at $x=1$

$ \begin{array}{r} \Rightarrow \quad \mathop {\lim }\limits_{x \to {1^ - }} f(x)=f(1)=1 \\ =\mathop {\lim }\limits_{x \to {1^ + }} f(x) \end{array} $

Now,

$\mathop {\lim }\limits_{x \to {1^ - }} f(x)=\mathop {\lim }\limits_{x \to {1^ - }} b-\left\{\frac{\sin [x-1]-[x-1]}{[x-1]^3}\right\} $

Put $x-1=t$

$ \begin{aligned} & =\mathop {\lim }\limits_{x \to {1^ - }} b-\left\{\frac{\sin [t]-[t]}{[t]^3}\right\} \\ & =b-\left\{\frac{\sin (-1)-(-1)}{(-1)^3}\right\} \end{aligned} $

$ =b+(-\sin 1+1)=b-\sin 1+1 $

and

$ \mathop {\lim }\limits_{x \to {1^ +}} f(x)=\mathop {\lim }\limits_{x \to {1^+ }}\left(a-\frac{\sin [x-1]}{x-1}\right) $

Put $x-1=t$

$ =\mathop {\lim }\limits_{x \to {1^ + }}\left(a-\frac{\sin [t]}{t}\right)=a-1 $

By Eq. (i), $a=2$ and $b-\operatorname{sinl}+1=1$

$ \begin{array}{ll} \Rightarrow & a=2 \text { and } b=\sin 1 \\ \Rightarrow & a+b=2+\sin 1 \end{array} $

$\mathop {\lim }\limits_{y \to 0} \frac{\left(1+\sqrt{1+y^4}\right)^{\frac{1}{2}}-\sqrt{2}}{y^4}$

Using L'Hospital's rule,

$ \begin{aligned} & \lim _{y \rightarrow 0} \frac{\frac{1}{2}\left[1+\sqrt{1+y^4}\right]^{\frac{-1}{2}}\left(\frac{1}{2}\left(1+y^4\right)^{\frac{-1}{2}} 4 y^3\right)}{4 y^3} \\ & \Rightarrow \frac{1}{2} \frac{1}{(\sqrt{2})}\left(\frac{1}{2}\right)=\frac{1}{4 \sqrt{2}} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{\cos 2 x-\cos 4 x}{1-\cos 2 x} \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{2 \sin 3 x \sin x}{2 \sin ^2 x}=k \end{aligned} \end{aligned} $

$ \begin{aligned} &\lim _{x \rightarrow 0} \frac{3 \sin x-4 \sin ^3 x}{\sin x}=3=k\\ &\text { Now, } \lim _{x \rightarrow 3}\left(\frac{x^3-27}{x^4-81}\right) \Rightarrow \lim _{x \rightarrow 3} \frac{3 x^2}{4 x^3}\\ &=\frac{3(3)^2}{4(3)^3}=\frac{1}{4} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=\left\{\begin{array}{cc} 1+\cos x, & x \leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{array}\right. \\ & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ & 2=a \Rightarrow a=2 \\ & \lim _{x \rightarrow 2^{-}} f(x)=f(2)=\lim _{x \rightarrow 2^{+}} f(x) \\ & \Rightarrow \quad a-2=4-b^2 \Rightarrow b^2=4 \\ & a^2+b^2=8 \end{aligned} \end{aligned} $

$\mathop {\lim }\limits_{x \to - \infty } \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}$

$ \therefore \quad|x|=-x $

Now, dividing $N^t$ and $D^t$ by $x^3$, we get

$ \mathop {\lim }\limits_{x \to - \infty } \frac{5-\frac{\sin 5 x}{x}}{\frac{\cos 4 x}{x^2}-7+\frac{4}{x^2}+\frac{3}{x^3}} $

Clearly, $\mathop {\lim }\limits_{x \to - \infty } \frac{\sin 5 x}{x}=0$

and $\mathop {\lim }\limits_{x \to - \infty } \frac{\cos 4 x}{x^2}=0$

$ \begin{gathered} \therefore \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} \\ =\frac{5}{-7}=-\frac{5}{7} \end{gathered} $

$\mathop {\lim }\limits_{x \to {a^ + }} f(x)=p, \mathop {\lim }\limits_{x \to {a^ - }} f(x)=m$

and $\quad f(a)=k$

$ \begin{array}{lc} & \mathrm{LHL}=\mathrm{RHL}=f(a) \\ \therefore & p=m=k \end{array} $

$p-m=0$ and $p-k=0$, then $f(x)$ is right continuous at $x=a$.

$ \begin{aligned} &\text { We have, } f(0)=a\\ &\begin{aligned} & \text { LHL }=\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2} \times \frac{16}{16} \\ &=\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{(4 x)^2} \times 16=\frac{1}{2} \times 16=8 \\ & \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4}} \\ &=\lim _{x \rightarrow 0} \frac{\sqrt{x}[\sqrt{16+\sqrt{x}}+4]}{16+\sqrt{x}-16} \\ &=\lim _{x \rightarrow 0} \sqrt{16+\sqrt{x}}+4=8 \\ & \therefore \quad a=8 \end{aligned} \end{aligned} $

$\mathop {\lim }\limits_{x \to \infty } \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2 x-4}}$

By rationalizing $N^r$ and $D^r$, we get

$ \begin{aligned} \lim _{x \rightarrow 0} & \frac{2-1-\cos x}{\sqrt{2}+\sqrt{1+\cos x}} \times \frac{\sqrt{15+\cos 2 x}+4}{15+\cos 2 x-16} \\ & =\lim _{x \rightarrow 0} \frac{1-\cos x}{-(1-\cos 2 x)} \times \frac{4+4}{\sqrt{2}+\sqrt{2}} \\ & =-\frac{8}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{\frac{1-\cos x}{x^2} \times x^2}{1-\frac{\cos 2 x}{(2 x)^2} \times 4 x^2} \end{aligned} $

Using result $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}$

$ =-\frac{8}{2 \sqrt{2}} \times \frac{\frac{1}{2}}{\frac{1}{2}} \times \frac{1}{4}=-\frac{1}{\sqrt{2}} $