Limits, Continuity and Differentiability

$ \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 $)$

$ \lim \limits_{x \rightarrow \frac{3}{2}} \frac{2 x(2 x-3)\left(4 x^{2}+6 x+9\right)}{(2 x)^{1 / 3}-3^{1 / 3}} $

$ \lim \limits_{x \rightarrow \frac{3}{2}} 2 x\left|\frac{8 x^{3}-27}{(2 x)^{1 / 3}-3^{1 / 3}}\right| $

$ 3 \lim \limits_{x \rightarrow \frac{3}{2}} \frac{24 x^{2}}{\frac{1}{3}(2 x)^{-\frac{2}{3}} \cdot 2}=\frac{72 \times 3\left(\frac{3}{2}\right)^{2}}{3^{-\frac{2}{3}}} $

$ =72 \times 3 \times \frac{9}{4} \times 3^{\frac{2}{3}} \times \frac{1}{2} $

$ =\frac{1}{2} \times 18 \times 27 \times 3^{2 / 3}=9 \times 27 \times 3^{2 / 3} $

$ =3^{2+3+\frac{2}{3}}=3^{\frac{17}{3}}=\sqrt[3]{3^{17}} $

To evaluate the limit of the function $ f(x) $ as $ x $ approaches 0, we start with the given expression and simplify step-by-step:

$ \lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{\left( 4^x - 1 \right)^4 \cot(x \ln 4)}{\sin(x \ln 4) \ln\left(1 + x^2 \ln 4\right)} $

First, rewrite the terms using a convenient limit form:

$ = \lim\limits_{x \to 0} \frac{\left(\frac{4^x - 1}{x}\right)^4 x^4 \cos(x \ln 4)}{\left(\frac{\sin(x \ln 4)}{x \ln 4}\right)^2 x^2 (\ln 4)^2} \times \frac{\ln\left(1 + x^2 \ln 4\right)}{x^2 \ln 4} \times x^2 \ln 4 $

Here, consider the limit:

$\lim\limits_{x \to 0} \frac{4^x - 1}{x} = \ln 4$

$\lim\limits_{x \to 0} \frac{\sin(x \ln 4)}{x \ln 4} = 1$

$\lim\limits_{x \to 0} \frac{\ln(1 + x^2 \ln 4)}{x^2 \ln 4} = 1$

Substituting these values into the expression:

$ = \lim\limits_{x \to 0} \frac{(\ln 4)^4}{(\ln 4)^3} = \ln 4 = k $

Thus, the value of $ e^k $ is:

$ e^k = e^{\ln 4} = 4 $

If $ x \in\left[0, \frac{\pi}{3}\right) \cos x > \frac{1}{2} $

then, $ \lim _{x \rightarrow a}=\frac{(2 \cos x-1)}{2 \cos x-1}=1 $

The function $ f(x) = \frac{|x-a|}{x-a} $ can be analyzed by considering different cases for the value of $ x $ relative to $ a $.

For $ x > a $, the expression $ |x-a| $ simplifies to $ x-a $. Thus, the function becomes:

$ f(x) = \frac{x-a}{x-a} = 1 $

For $ x < a $, the expression $ |x-a| $ simplifies to $-(x-a)$, which means the function is:

$ f(x) = \frac{-(x-a)}{x-a} = -1 $

Thus, the function $ f(x) $ can be rewritten as a piecewise function:

$ f(x) = \begin{cases} 1, & \text{if } x > a \\ -1, & \text{if } x < a \end{cases} $

Based on this function:

$ f(x) $ is a constant function (equal to 1) when $ x > a $.

It is discontinuous at $ x = a $ because the limits from the left and the right of $ a $ do not match. Specifically, the left-hand limit approaches $-1$ while the right-hand limit approaches $1$, and thus the function does not have a limit at this point.

In summary, $ f(x) $ is not continuous at $ x = a $ and is constant when $ x > a $.

Given the function:

$ f(x) = 3x^{15} - 5x^{10} + 7x^{5} + 50 \cos(x-1) $

We need to evaluate the limit:

$ \lim\limits_{h \rightarrow 0} \frac{f(1-h) - f(1)}{h^3 + 3h} $

This expression is initially an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's rule.

Using L'Hôpital's rule, the limit becomes:

$ \lim\limits_{h \rightarrow 0} \frac{-f'(1-h)}{3h^2 + 3} $

We then evaluate this limit by substituting $h = 0$:

$ \frac{-f'(1)}{3} $

To find $f'(x)$, we differentiate $f(x)$:

$ f'(x) = 45x^{14} - 50x^{9} + 35x^{4} - 50 \sin(x-1) $

Substitute $x = 1$ into $f'(x)$:

$ f'(1) = 45(1)^{14} - 50(1)^{9} + 35(1)^{4} - 50 \sin(0) $

$ f'(1) = 45 - 50 + 35 + 0 $

$ f'(1) = 30 $

Substitute back into the limit expression:

$ \frac{-30}{3} = -10 $

Thus, the value of the limit is $-10$.

To ensure the function $ f(x) $ is differentiable at $ x = 0 $, it must also be continuous at this point. Therefore, we need:

$ \lim\limits_{x \to 0} \frac{(e^{kx} - 1) \sin kx}{4 \tan x} = P $

Let's simplify the limit step-by-step:

Convert the expression using known limits:

$ \lim\limits_{x \to 0} \frac{kx}{4} \left[\frac{e^{kx} - 1}{kx}\right] \left[\frac{\sin kx}{kx}\right] \cdot \frac{kx}{x \left(\frac{\tan x}{x}\right)^2} $

Recognize standard limits for small $ x $:

$\lim\limits_{x \to 0} \frac{e^{kx} - 1}{kx} = 1$

$\lim\limits_{x \to 0} \frac{\sin kx}{kx} = 1$

$\lim\limits_{x \to 0} \frac{\tan x}{x} = 1$

Plug these into the equation:

$ \lim\limits_{x \to 0} \frac{k^2 x^2}{4x} = \lim\limits_{x \to 0} \frac{k^2 x}{4} = P $

As $ x \to 0 $, the expression $ \frac{k^2 x}{4} \to 0 $, so $ P = 0 $.

Since $ P = 0 $, let's find the derivative at $ x = 0 $:

$ f'(x) = \lim\limits_{h \to 0} \frac{f(h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{\frac{(e^{kh} - 1) \sin kh}{4 \tan h}}{h} $

Applying the same logic and limits as before, the derivative becomes:

$ f'(0) = \frac{k^2}{4} $

Therefore, $ P = 0 $ and $ f'(0) = \frac{k^2}{4} $.

To apply Rolle's Theorem, we need the function to satisfy $ f(a) = f(b) $ while $ a \neq b $. For this function, consider :

$ \lim\limits_{x \rightarrow 0} x^{p} \log x = 0 $

Expanding the logarithm using Taylor series, we get :

$ \lim\limits_{x \rightarrow 0} x^{p} \left[x - 1 - \frac{(x-1)^{2}}{2} + \frac{(x-1)^{3}}{3} + \ldots\right] = 0 $

If $ p = 0 $, the limit does not approach zero.

Therefore, the minimum value of $ p $ that satisfies the condition for the limit to be zero is $ p = 1 $.

Given, $ \lim \limits_{x \rightarrow 4} \frac{2 x^{2}+(3+2 a) x+3 a}{x^{3}-2 x^{2}-23 x+60}=\frac{11}{9} $

Put $ x=4 $

$ =\frac{4}{32+12+8 a+3 a} \frac{64-32-92+60}{64+11 a} $

$ \because $ Numerator must be zero

$ \Rightarrow \quad a=\frac{-44}{11}=-4 $

Then, $ \lim \limits_{x \rightarrow-4} \frac{x^{2}+9 x+20}{x^{2}-x-20} $

$ \lim \limits_{x \rightarrow-4} \frac{(x+5)(x+4)}{(x-5)(x+4)} \Rightarrow \frac{-4+5}{-4-5}=\frac{-1}{9} $

$ \because f(x) $ is continuous at $ x=4 $.

$ \begin{aligned} f(4) & =\frac{64-125}{16-25}=\frac{61}{9} \\\\ f\left(4^{+}\right) & =\frac{b^{4}-1}{4}=\frac{61}{9} \\\\ b^{4} & =\frac{244+9}{9}=\frac{253}{9} \Rightarrow 9 b^{4}=253 \end{aligned} $

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

So, $ f\left(1^{+}\right)=f(1)=\frac{1-125}{1-25}=\frac{31}{6} \Rightarrow f\left(1^{-}\right)=a $

$ \Rightarrow 6 a+9 b^{4} \Rightarrow 31+253 \Rightarrow 284 $

$ \begin{aligned} & \lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4} \\\\ & =\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \sin ^4 \theta+4 \sin ^2 \theta \cos ^2 \theta+5 \cos ^4 \theta}{(3 \cos \theta-2 \sin \theta)^4} \\\\ & =\frac{8(1)^4+4(1)^2(0)^2+5(0)^4}{(3(0)-2(1))^4}=\frac{8}{16}=\frac{1}{2} \end{aligned} $

Given,

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

$ \text { is continuous at } x=0 $

$ \begin{array}{lr} \therefore & \lim _{x \rightarrow 0^{-}} f(x)=f(0) \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2}=a \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} \frac{1-\left(1-2 \sin ^2 2 x\right)}{x^2}=a \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 2 x}{x^2}=a \\\\ \Rightarrow & \lim _{x \rightarrow 0^{-}} 8\left(\frac{\sin 2 x}{2 x}\right)^2=a \\\\ \Rightarrow & 8(1)^2=a \Rightarrow a=8 \end{array} $

We have,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x} \,\,\,\,\,\,\,\,\,\,\,\,\left(\frac{0}{0}\right. form ) \\\\ = & \lim _{x \rightarrow 0} \frac{\ln 3\left(3^{\sin x} \cdot \cos x\right)-\ln 2\left(2^{\operatorname{an} x} \cdot \sec ^2 x\right)}{\cos x} \\\\ = & \frac{\ln 3 \cdot 1-\ln 2 \cdot 1}{1}=\ln 3-\ln 2=\ln \left(\frac{3}{2}\right) \end{aligned} $

We have,

$ f(x)=\left\{\begin{array}{cc} \frac{\cos a x-\cos 9 x}{x^2} & , \text { if } x \neq 0 \\\\ 16 & \text {, if } x=0 \end{array}\right. $

Now, $\lim _{x \rightarrow 0} f(x)=16 \quad \ldots \text { (ii) }$

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

$ \lim _{x \rightarrow 0} \frac{\cos a x-\cos 9 x}{x^2}=16 $

$ \begin{aligned} &\text { Take LHS }\\\\ &\begin{aligned} & \text { HS } \\\\ & \begin{array}{l} \lim _{x \rightarrow 0} \frac{\cos a x-\cos 9 x}{x^2} \\\\ =\lim _{x \rightarrow 0} \frac{-a \sin a x+9 \sin 9 x}{2 x} \\\\ \text { [using L.' Hospital's rule] } \\\\ =\lim _{x \rightarrow 0} \frac{-a^2 \cos a x+81 \cos 9 x}{2} \\\\ =\frac{81-a^2}{2}=16 \quad \text { [from, Eq. (i)] } \\\\ 81-a^2=32 \Rightarrow a^2=49 \Rightarrow a= \pm 7 \end{array} \end{aligned} \end{aligned} $

We have,

$ f(x)=\left\{\begin{array}{ll} \frac{8}{x^{3}}-6 x, & \text { if } 0 < x \leq 1 \\\\ \frac{x-1}{\sqrt{x}-1}, & \text { if } x > 1 \end{array}\right. $

$ \begin{array}{l} f(1)=2 \text { and } \\\\ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(\frac{8}{x^{3}}-6 x\right)=2=\text { LHL } \end{array} $

$ \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \frac{x-1}{\sqrt{x}-1} $

$ =\lim _{x \rightarrow 1^{+}} \frac{1-0}{\frac{1}{2 \sqrt{x}}-0}=\frac{21}{1}=2 $

$ \because \quad f(\mathrm{l})=\mathrm{LHL}=\text { RHL } $

$ \Rightarrow \quad f(x) $ is continuous at $ x=1 $

And for differentiability at $ x=1 $,

$ \begin{array}{l} \text { LHD }=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\\\ =\lim _{h \rightarrow 0} \frac{\frac{8}{(1-h)^{3}}-6(1-h)-f(2)}{-h} \\\\ =\lim _{h \rightarrow 0} \frac{8-6(1-h)^{4}-2(1-h)^{3}}{-h(1-h)^{3}} \end{array} $

$ \begin{array}{l}=\lim _{h \rightarrow 0} \frac{0+24(1-h)^{3}+6(1-h)^{2}}{-\left[(1-h)^{3}-3 h(1-h)^{2}\right]} \\\\ =-\lim _{h \rightarrow 0} \frac{24(1-h)+6}{1-h-3 h}=-30 \\\\ \text { RHD }=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \\\\ =\lim _{h \rightarrow 0} \frac{\frac{1+h-1}{\sqrt{1+h}-1}-2}{h} \\\\ =\lim _{h \rightarrow 0} \frac{h-2 \sqrt{1+h}+2}{(\sqrt{1+h}-1) h}\end{array} $

$ \begin{array}{l} =\lim _{h \rightarrow 0} \frac{1-\frac{1}{\sqrt{1+h}}}{\frac{1}{2 \sqrt{1+h}} \cdot h+\sqrt{1+h}-1} \\\\ =2 \lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h+2(1+h)-2 \sqrt{1+h}} \\\\ =2 \lim _{h \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+h}}}{1+2-\frac{1}{\sqrt{1+h}}}=2 \times \frac{\frac{1}{2}}{3-1}=\frac{1}{2} \end{array} $

$ \because \mathrm{LHD} \neq \mathrm{RHD} $

Hence, at $ x=1, f(x) $ is continuous but not differentiable.

$ \begin{aligned} & \text L=\lim _{n \rightarrow \infty}\left[\begin{array}{c} \left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right) \\\\ \left(1+\frac{9}{n^2}\right) \ldots(2) \end{array}\right]^{1 / n} \\\\ & =\lim _{n \rightarrow \infty}\left(\prod_{K=1}^n\left(1+\frac{K^2}{n^2}\right)\right)^{1 / n} \\\\ & L=\lim _{n \rightarrow \infty} e^{\left(\frac{1}{n} \sum_{K=1}^n \ln \left(1+\frac{K^2}{n^2}\right)\right)} \quad \ldots \text { (ii) } \end{aligned} $

Sum can be approximated by an integral in the limit as $n \rightarrow \infty$

$ \Rightarrow \frac{1}{n} \sum_{K=1}^n \ln \left(1+\frac{K^2}{n^2}\right) \approx \int_0^1 \ln \left(1+x^2\right) d x $

Using integration by parts

$ \begin{aligned} \int_0^1 \ln \left(1+x^2\right) d x=\left[\ln \left(1+x^2\right)\right. & \cdot x]_0^1 \\\\ & -\int \frac{2 x^2+2-2}{1+x^2} d x \end{aligned} $

$ =\ln 2-\int_0^1 2 d x+2 \int_0^1 \frac{1}{1+x^2} d x=\ln 2-2+\frac{\pi}{2} $

$ \begin{aligned} & \text { So, from Eq. (i), we get } \\\\ & L=e^{\ln 2-2+\frac{\pi}{2}}=2^{\frac{\pi}{2}-2}=2^{\left(\frac{\pi-4}{2}\right)} \end{aligned} $

$ \begin{aligned} &\text { Using rationalisation }\\ &\begin{aligned} & \frac{(2 \tan x+\cos x-1+x)}{\left(\sqrt{4 \sin ^2 x+2 \tan x+1}\right.} \\ = & \lim _{x \rightarrow 0} \frac{\left.+\sqrt{3 \tan ^2 x+\sin x+1}\right)}{4 \sin ^2 x+2 \tan x-3 \tan ^2 x-\sin x} \\ = & \lim _{x \rightarrow 0} \frac{x\left[\frac{2 \tan x}{x}-\left(\frac{1-\cos x}{x^2}\right) x+1\right][\sqrt{1}+\sqrt{1}]}{\sin x(4 \sin x+2 \sec x-3 \sec x \tan x-1)} \\ = & \frac{(2-0+1)(2)}{(0+2-0-1)}=6 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)} \\ & x=\frac{\pi}{4}+h \\ & =\lim _{h \rightarrow 0} \frac{1-\tan ^4\left(\frac{\pi}{4}+h\right)}{\tan ^3\left(\frac{\pi}{4}+h\right)(-\sin h)} \\ & =\lim _{h \rightarrow 0} \frac{\left(1+\tan ^2\left(\frac{\pi}{4}+h\right)\right.}{\tan ^3\left(\frac{\pi}{4}+h\right)} \frac{\left(1-\tan ^2\left(\frac{\pi}{4}+h\right)\right)}{-\sin h} \\ & =\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}+h\right)}{-\sin h} \cdot\left[1+\tan \left(\frac{\pi}{4}+h\right)\right] \times 2 \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}+h\right)}{\left[1+\tan \left(\frac{\pi}{4}+h\right)\right]} \frac{\left[1+\tan \left(\frac{\pi}{4}+h\right)\right]^2}{[-\sin h]} \times 2 \\ & =\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}-h\right)}{-\sin h} \times\left(2^2 \times 2\right. \\ & =\lim _{h \rightarrow 0} \frac{\tan (h)}{h} \frac{h}{\sin h} \times 8=8 \end{aligned} $

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

The quadratic equation whose roots are $l, m$, where

$ \begin{aligned} l & =\lim _{\theta \rightarrow 0} \frac{3 \sin \theta-4 \sin ^3 \theta}{\theta} \\ & =\lim _{\theta \rightarrow 0} \frac{3 \sin \theta}{\theta}-4 \frac{\sin ^3 \theta}{\theta}=3 \\ l & =3 \\ m & =\lim _{\theta \rightarrow 0} 2 \frac{\tan \theta}{\theta\left(1-\tan ^2 \theta\right)} \\ & =\lim _{\theta \rightarrow 0} 2 \frac{\tan \theta}{\theta} \cdot \frac{1}{\left(1-\tan ^2 \theta\right)}=2 \\ m & =2 \end{aligned} $

Thus, the equation is $x^2-5 x+6=0$

Given, $\mathop {\lim }\limits_{x \to 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}\left[\frac{0}{0}\right.$ form $]$ On applying L'Hospital's rule, we get

$ \begin{aligned} & =\lim _{x \rightarrow 2} \frac{\frac{1}{3}(6+x)^{-2 / 3}+\frac{1}{3}(10-x)^{-2 / 3}}{1} \\ & =\frac{1}{3}\left[(8)^{-2 / 3}+(8)^{-2 / 3}\right]=\frac{1}{3}\left[2(8)^{-2 / 3}\right] \\ & =\frac{1}{3}\left[2\left(2^3\right)^{-2 / 3}\right]=\frac{1}{3} \times 2 \times(2)^{-2} \\ & =\frac{1}{3} \times 2 \times \frac{1}{4}=\frac{1}{6} \end{aligned} $

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

$ \text { } \begin{aligned} & \lim _{x \rightarrow-2^{+}}\left([x]^2-[x]-2\right)+\lim _{x \rightarrow-3^{-}}\left([x]^2-4[x]+3\right) \\ &=\left((-2)^2-(-2)-2\right)+\left((-4)^2-4(-4)+3\right) \\ &=(4+2-2)+(16+16+3)=39 \end{aligned} $

$ \begin{aligned} &\text { } \lim _{x \rightarrow 0} \frac{\left(3^{2 x}-\sqrt{x+1}\right) \sin 5 x}{1-\cos 4 x}\\ &=\lim _{x \rightarrow 0} \frac{3^{2 x}-\sqrt{x+1}}{x} \times \frac{\sin 5 x}{5 x} \times \frac{5 \cdot(2 x)^2}{2 \sin ^2 2 x} \times \frac{1}{4} \end{aligned} $

$ =\frac{5}{8}\left[\begin{array}{r} \lim _{x \rightarrow 0}\left(\frac{3^{2 x}-1}{x}+\frac{1-\sqrt{x+1}}{x}\right) \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \\ \lim _{x \rightarrow 0}\left(\frac{2 x}{\sin 2 x}\right)^2 \end{array}\right] $

$ \begin{aligned} & =\frac{5}{8}\left[2 \log _e 3-\frac{1}{2}\right]=\frac{5}{16}\left(\log _e 3^4-1\right) \\ & =\frac{5}{16} \log _e\left(\frac{81}{e}\right) \end{aligned} $

$\lim\limits_{x \rightarrow 1}(1-x) \tan \left(\frac{\pi}{2} x\right)$

Let $1-x=t \Rightarrow x=1-t$

So, as $\Rightarrow t \rightarrow 0$

$ \begin{aligned} &Now,\,\,\, \lim _{t \rightarrow 0} t \tan \left(\frac{\pi}{2}(1-t)\right) \\ & =\lim _{t \rightarrow 0} t \tan \left(\frac{\pi}{2}-\frac{\pi t}{2}\right)=\lim _{t \rightarrow 0} t \cot \left(\frac{\pi t}{2}\right) \\ & =\lim _{t \rightarrow 0} \frac{t}{\tan \left(\frac{\pi}{2} t\right)}=\frac{2}{\pi} \lim _{t \rightarrow 0}\left(\frac{\frac{\pi}{2} t}{\tan \frac{\pi t}{2}}\right) \\ & =\frac{2}{\pi} \cdot 1=\frac{2}{\pi} \end{aligned} $

Given, $f(9)=9$ and $f^{\prime}(9)=4$ and we have to find

$ \lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} $

$ \begin{aligned} &\text { Apply } L^{\prime} \text { Hospital Rule, we get }\\ &\begin{aligned} \lim _{x \rightarrow 9} \frac{\frac{1}{2 \sqrt{f(x)}} \cdot f^{\prime}(x)}{\frac{1}{2 \sqrt{x}}} & =\frac{\frac{1}{\sqrt{f(9)}} f^{\prime}(9)}{\frac{1}{\sqrt{9}}} \\ & =\frac{4 / 3}{1 / 3}=4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\ & =\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{\left(2 x^2+x-3\right)(\sqrt{x}+1)} \\ & =\lim _{x \rightarrow 1} \frac{(2 x-3)(x-1)}{(2 x+3)(x-1)(\sqrt{x}+1)} \\ & =\frac{(2-3)}{(2+3)} \times \frac{1}{(\sqrt{1}+1)}=-\frac{1}{5} \times \frac{1}{2}=\frac{-1}{10} \end{aligned} $

To solve the given problem, we need to evaluate the limit:

$ \lim_{x \rightarrow \infty} x\left(a^{1/x} + b^{1/x} + c^{1/x} - 3k^{1/x}\right) = 0 $

We start by substituting $x = \frac{1}{y}$, which implies $y \rightarrow 0$ as $x \rightarrow \infty$. The expression becomes:

$ \lim_{y \rightarrow 0} \frac{a^y + b^y + c^y - 3k^y}{y} = 0 $

Using the known limit:

$ \lim_{y \rightarrow 0} \frac{a^y - 1}{y} = \log_e a $

This allows us to rewrite the expression as:

$ \lim_{y \rightarrow 0}\left[\frac{a^y - 1}{y} + \frac{b^y - 1}{y} + \frac{c^y - 1}{y} - 3 \frac{k^y - 1}{y}\right] = 0 $

This simplifies to:

$ \log_e a + \log_e b + \log_e c - 3 \log_e k = 0 $

Rewriting it gives:

$ \log_e \left(\frac{abc}{k^3}\right) = 0 $

Thus, we have:

$ \frac{abc}{k^3} = 1 $

Solving for $k$, we find:

$ k = (abc)^{1/3} $

$ \text { } \mathop {\lim }\limits_{x \to 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2} $

$ \begin{aligned} & \therefore \lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2} \\ & =\lim _{x \rightarrow 2} \frac{(x-2)\left(x^2+x+1\right)}{(x-2)\left(2 x^2+x-1\right)} \\ & =\lim _{x \rightarrow 2} \frac{x^2+x+1}{2 x^2+x-1}=\frac{4+2+1}{8+2-1}=\frac{7}{9} \end{aligned} $

$ \begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)} +\cos (2020 x)] \\ & =\lim _{x \rightarrow 0} 4\left[\frac{\sin 2022 x}{x}-\frac{\sin 2020 x}{x}\right] \\ & \times\left[\frac{1}{\cos 2022 x+2 \cos 2021 x+\cos 2020 x}\right] \end{aligned} $

$ =\mathop {\lim }\limits_{x \to 0} 4\left[\begin{array}{l} 2022 \cdot\left(\frac{\sin 2022 x}{2022 x}\right) -2020\left(\frac{\sin 2020 x}{2020 x}\right) \end{array}\right] $

$ \begin{aligned} & \times\left[\frac{1}{\cos 2022 x+2 \cos 2021 x+\cos 2020 x}\right] \\ &= 4[2022 \times 1-2020 \times 1] \\ & \times\left[\frac{1}{\cos 0+2 \cos 0+\cos 0}\right] \\ &= 4(2022-2020) \times\left[\frac{1}{1+2+1}\right] \\ &= 4 \times 2 \times \frac{1}{4}=2 \end{aligned} $

Given, $f(x)=2 x-[2 x]$

$[x] \rightarrow$ Greatest Integer Function

Now,

$ \begin{aligned} & l_1=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h) \\ & =\lim _{h \rightarrow 0} 2(2-h)=\lim _{h \rightarrow 0} 4-2 h-3=1 \end{aligned} $

Now, $l_2=\mathop {\lim }\limits_{x \to {2^ + }} f(x)=\mathop {\lim }\limits_{h \to {0 }} f(2+h)$

$ \begin{aligned} & =\lim _{h \rightarrow 0} 2(2+h)-[4+2 h] \\ & =\lim _{h \rightarrow 0}(4+2 h-4)=0 \\ \therefore l_1+l_2 & =0+1=1 \end{aligned} $

$\mathop {\lim }\limits_{x \to 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}\left[\frac{0}{0}\right.$ form $]$

$ =\frac{1}{2} \frac{\mathop {\lim }\limits_{x \to 0} \frac{2^x-1}{x}}{\mathop {\lim }\limits_{x \to 0} \frac{\log (1+2 x)}{2 x}} \cdot e^{\mathop {\lim }\limits_{x \to 0} (1+\sin x-1) \times \frac{2}{\sin x}} $

Using $L^{\prime}$ hospital rule,

$ \begin{aligned} & \frac{1}{2} \frac{\mathop {\lim }\limits_{x \to 0} 2^x \log 2}{\mathop {\lim }\limits_{x \to 0} \frac{1}{1+2 x}}, e^{\mathop {\lim }\limits_{x \to 0} \sin x \times \frac{2}{\sin x}} \\ & =\frac{1}{2} \frac{\log 2}{1} e^2=e^2\left(\frac{1}{2} \log 2\right)=e^2 \log \sqrt{2} \end{aligned} $

Given $f(x)$ is a differentiable function, and

$ f(0)=0, f^{\prime}(0)=20, x \in\left(0, \frac{\pi}{2}\right] $

$ \begin{aligned} &\begin{aligned}& A(x)=2 f(x) \operatorname{cosec}(4 x) +4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x &\\ &\lim _{x \rightarrow 0} A(x) =\lim _{x \rightarrow 0} \frac{2 f(x)}{\sin 4 x} +4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x \end{aligned}\\ &\text { Using L'Hospital rule }\\ &\begin{aligned} & \lim _{x \rightarrow 0} A(x)=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)}{4 \cos 4 x} +4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x \\ & =\frac{2 f^{\prime}(0)}{4 \cos 0}+4 f(0)\left(\cos ^2 0+1\right)-4 \cos ^2 0 \\ & =\frac{2 \times 20}{4 \times 1}+0-4 \times 1=10-4=6 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } x=\log _e\left(\cot \left(\frac{\pi}{4}+\theta\right)\right) \\ & =\lim _{\theta \rightarrow 0} \frac{\theta}{\sinh x \cosh x} \\ & \begin{array}{l} \lim _{\theta \rightarrow 0} \frac{\theta}{\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^x+e^{-x}}{2}\right)} \\ =\lim _{\theta \rightarrow 0} \frac{4 \theta}{e^{2 x}-e^{-2 x}} \\ =\lim _{\theta \rightarrow 0} \frac{4 \theta}{\cot ^2\left(\frac{\pi}{4}+\theta\right)-\tan ^2\left(\frac{\pi}{4}+\theta\right)} \end{array} \end{aligned} $

$ \begin{aligned} &\begin{array}{r} =\lim _{\theta \rightarrow 0} \frac{4}{-2 \cot \left(\frac{\pi}{4}+\theta\right) \operatorname{cosec}^2\left(\frac{\pi}{4}+\theta\right)} \\ -2 \tan \left(\frac{\pi}{4}+\theta\right) \sec ^2\left(\frac{\pi}{4}+\theta\right) \end{array}\\ &\text { [ ∵ Apply L' Hospital rule] } \end{aligned} $

$ \begin{aligned} & =\frac{4}{-2(1)(2)-2(1)(2)}=\frac{4}{-4-4} \\ & =-\frac{4}{8}=-\frac{1}{2} \end{aligned} $

$ \begin{aligned} \\ \begin{array}{r} \text { } \mathop {\lim }\limits_{x \to 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)\right. \left.+\frac{x^2-4}{x^3-2 x-4}\right] \\ =\mathop {\lim }\limits_{x \to 2}\left[(x-2)^2 \cos \left(\frac{2}{x-2}\right)\right. \left.+\frac{(x-2)(x+2)}{x^3-2 x^2+2 x^2-4 x+2 x-4}\right] \\ =\mathop {\lim }\limits_{x \to 2}\left[(x-2)^2 \cos \left(\frac{2}{x-2}\right)\right] +\frac{(x-2)(x+2)}{(x-2)\left(x^2+2 x+2\right)} \\ =\mathop {\lim }\limits_{x \to 2}\left[(x-2)^2 \cos \left(\frac{2}{x-2}\right)\right] +\mathop {\lim }\limits_{x \to 2} \frac{(x+2)}{\left(x^2+2 x+2\right)} \,...(i)\end{array} \end{aligned} $

$ \cos \left(\frac{2}{x-2}\right) \in[-1,1] $

So, $\mathop {\lim }\limits_{x \to 2}(x-2)^2 \cos \left(\frac{2}{x-2}\right)=0$

Now, substitute $x=2$ in Eq. (i),

$ 0+\frac{4}{10}=\frac{2}{5} $