Limits, Continuity and Differentiability

Given that

$ f(x)=\left\{\begin{array}{cc} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6} & \text { for } x \neq 3 \\ l & \text { for } x=3 \end{array}\right. $

is continuous at $x=3$

$ \begin{aligned} & \Rightarrow \text { RHL }=\text { LHL }=\text { Value of function } \\ & \therefore \lim _{x \rightarrow 3^{+}} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}=\text { Value of } \end{aligned} $

function at $x=3$

$ \begin{aligned} & \Rightarrow \lim _{x \rightarrow 3^{+}} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}=l \\ & \Rightarrow \lim _{x \rightarrow 3^{+}} \frac{2 x^2+(k+2) x+9}{(3 x+2)(x-3)}=l \end{aligned} $

For limit to be exist, numerator also have a factor $(x-3)$ so

$ 2 x^2+(k+2) x+9=(x-3)(2 x+a) $

Expand and equate the coefficients

$ \begin{aligned} & k+2=a-6 \\ & 9=-3 a \Rightarrow a=-3 \Rightarrow k=-11 \end{aligned} $

So, $2 x^2+(k+2) x+9=(x-3)(2 x-3)$

Then, $\lim \limits_{x \rightarrow 3^{+}} \frac{(x-3)(2 x-3)}{(3 x+2)(x-3)}=1$

$ \Rightarrow \quad \frac{6-3}{9+2}=l \Rightarrow l=\frac{3}{11} $

So, $l-k=\frac{3}{11}+11=\frac{124}{11}$

$\mathop {\lim }\limits_{x \to o} \left[ {{1 \over x} - {1 \over {{e^x} - 1}}} \right] = \mathop {\lim }\limits_{x \to o} {{{e^x} - 1 - x} \over {x({e^x} - 1)}}$

$ \text { Using L'Hospital's rule } $

$=\mathop {\lim }\limits_{x \to o} {{{e^x} - 1} \over {{e^x} - 1 + x{e^x}}}$

$ \text { Again, using L'Hospital rule, we get } $

$ = \mathop {\lim }\limits_{x \to o} {{{e^x}} \over {{e^x} - {e^x} + x{e^x}}} = {{{e^o}} \over {2{e^o} + 0}} = {1 \over 2}.$

$f(x)= \begin{cases}0, & x=0 \\ 2-x, & \text { for } 0 < x < 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 < x < 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{cases}$

continuity at $x=1$

As $x$ approaches 1 from the left

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

The value of $f(x)$ at $x=1$

$ f(1)=2 $

LHL $\neq$ RHL

Therefore, $f(x)$ is not continuous at $x=1$

Continuity at $x=2$

As $x$ approaches 2 from the left (1 < x < 2)

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

The value of $f(x)$ at $x=2$

$ f(x)=-\frac{3}{2} $

The LHL and actual value of the function at $x=2$ are equal to RHL

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

$ \text { Therefore, } f(x) \text { is continuous at } x=2 $

$f(x)=\left(\frac{1+x}{1-x}\right)^{1 / x}$

We will calculate the limit of this expression as $x$ approaches 0 .

Let $L=\left(\frac{1+x}{1-x}\right)^{1 / x}$

$L=\lim _{x \rightarrow 0} \frac{1}{x} \ln \left(\frac{1+x}{1-x}\right)^{1 / x} \ldots(\mathrm{i})($ natural $\log$ )

Let $y=\log \frac{1+x}{1-x}$

$ =\log (1+x)-\log (1-x) $

Using Taylor's expansions for $\log (1+x)$ and $\log (1-x)$ around $x=0$

$ \log (1+x) \approx x $

$ \log (1-x) \approx-x $

Then, $y=x-(-x)$

$ y=2 x \Rightarrow \log \left(\frac{1+x}{1-x}\right)=2 x \ldots \text { Eq. } $

From Eq. (i), we get

$ \begin{aligned} \log L=\lim _{x \rightarrow 0} \frac{1}{x} & \ln \left(\frac{1+x}{1-x}\right) \\ & =\lim _{x \rightarrow 0} \frac{1}{x} \times 2 x \text { using } \\ \log L & =\lim _{x \rightarrow 0} 2 \\ \log L & =2 \\ f(x) & =L=e^2 \end{aligned} $

$\because f(x)$ is continuous at, $x=0, f(0)=e^2$

The function $ f(x) = |x - 24| $ is being evaluated for its continuity and differentiability on the interval $[0, 25]$.

Continuity

The function $ f(x) = |x - 24| $ is continuous over the interval $[0, 25]$. The absolute value function is continuous across its domain, including the point $ x = 24 $ within this interval.

Differentiability

The function $ f(x) = |x - 24| $ is differentiable on the interval $(0, 25)$ except at the point $ x = 24 $. At $ x = 24 $, the left-hand derivative is $ f'(24) = -1 $ and the right-hand derivative is $ f'(24) = 1 $. Since the left-hand and right-hand derivatives at $ x = 24 $ are not equal, the function is not differentiable at this point.

For the subintervals $ 0 < x < 24 $ and $ 24 < x \leq 25 $, the derivative of $ f(x) $ is well defined.

Conclusion

Thus, the function $ f(x) = |x - 24| $ is continuous on $[0, 25]$ but not differentiable on $[0, 25]$, specifically due to the point $ x = 24 $.

$ \begin{aligned} &\text { We can observe that, }\\ &\begin{aligned} & T_n=\frac{1}{\sqrt{n^2-r^2}} \\ & \Rightarrow \lim _{n \rightarrow \infty} \sum_{n=0}^{n-1} \frac{1}{\sqrt{n^2-r^2}}=\lim _{n \rightarrow \infty} \sum_{n=0}^{n-1}{1 \over {n\sqrt {1 - {{\left( {{r \over n}} \right)}^1}} }} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } \frac{r}{n}=x \\ & \frac{1}{n}=d x \\ & \text { limit } r=0 \Rightarrow x=0 \\ & r=n-1 \\ & \Rightarrow \quad x=1-\frac{1}{n} \\ & x=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because n=\infty)\\ & \Rightarrow \int_0^1 \frac{d x}{\sqrt{1-x^2}} \\ & =\left(\sin ^{-1}\right)_0^1 \\ & =\sin ^{-1}(\mathrm{l})-\sin ^{-1}(0) \\ & =\frac{\pi}{2}-0=\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}= & \lim _{x \rightarrow 0} \frac{\sin \left(\pi\left(1-\sin ^2 x\right)\right)}{x^2} \\ & {\left[\because \sin ^2 x+\cos ^2 x=1\right] } \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2}=\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\left(\pi \sin ^2 x\right)} \times \frac{\left(\pi \sin ^2 x\right)}{\left(x^2\right)} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\left(\pi \sin ^2 x\right)} \cdot \lim _{x \rightarrow 0} \frac{\pi \sin ^2 x}{x^2} \\ & =\pi \quad\left\{\because \lim _{x \rightarrow 0} \frac{\sin \theta}{\theta}=1\right\} \end{aligned} $

we have

$ \begin{aligned} & \lim _{x \rightarrow 1}\left(\frac{x+x^2+x^3+\ldots+x^n-n}{x-1}\right) \\ = & \lim _{x \rightarrow 1}\left[\frac{x+x^2+x^3+\ldots+x^n-n}{(x-1)}\right] \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 1}\left[\begin{array}{c} x+x^2+x^3+\ldots+x^n \\ -(1+1+\ldots+1 \ldots n \text { times }) \\ (x-1) \end{array}\right] \\ & =\lim _{x \rightarrow 1}\left[\frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right)+\ldots+\left(x^n-1\right)}{(x-1)}\right] \end{aligned} $

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

So, divide each term is square bracket by $(x-1)$ as given in denominator and applying the limit, we get

$\mathop {\lim }\limits_{x \to 1} \left[ {{{(x - 1) + ({x^2} - {1^2}) + ({x^3} - {1^3}) + ... + ({x^n} - {1^n})} \over {(x - 1)}}} \right]$

$ \begin{aligned} & =\frac{1(1)^0+2(1)^1+3(l)^2+4(1)^3+\ldots+n(1)^{n-1}}{} \\ & {\left[\text { Formula used, } \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]} \end{aligned} $

$ \begin{aligned} & =1+2+3+4+\ldots+n \\ & =\frac{n(n+1)}{2} \end{aligned} $

Given that the functions is continuous

$ \begin{aligned} & \Rightarrow \lim _{x \Rightarrow 0} f(x)=f(0) \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\frac{\sqrt{1+0}-1}{0}=\frac{0}{0} \\ & {\left[\text { which is } \frac{0}{0} \text { from }\right]} \end{aligned} $

So, we apply L-Hospital rule

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\sqrt{1+x}-1)}{\frac{d}{d x} x}=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \sqrt{1+x}}{1} \\ & =\lim _{x \rightarrow 0} \frac{1}{2 \sqrt{1+x}} \end{aligned} $

On applying the limit, we get

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{1}{2 \sqrt{1+h}}=\frac{1}{2} \\ \therefore &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ f(0)=\frac{1}{2} \end{aligned} $

$ \begin{aligned} & \text { If } f(x)=\frac{5 x \cdot \operatorname{cosec}(\sqrt{x})-1}{(x-2) \operatorname{cosec}(\sqrt{x})} \\ & \text { then, } \lim _{x \rightarrow \infty} f\left(x^2\right)=\text { ? } \\ & f\left(x^2\right)=\frac{5\left(x^2\right) \operatorname{cosec}\left(\sqrt{x^2}\right)-1}{\left(x^2-2\right) \operatorname{cosec}\left(\sqrt{x^2}\right)} \\ & \lim _{x \rightarrow \infty} f\left(x^2\right)=\lim _{x \rightarrow \infty} \frac{5 x^2 \operatorname{cosec} x-1}{\left(x^2-2\right) \operatorname{cosec} x} \\ & =\lim _{x \rightarrow \infty} \frac{x^2 \operatorname{cosec} x\left(5-\sin x / x^2\right)}{x^2\left(1-\frac{2}{x^2}\right) \operatorname{cosec} x} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \infty} \frac{\left[5-\frac{\sin x}{x^2}\right]}{\left(1-\frac{2}{x^2}\right)} \\ & \text { Here, } \frac{\sin x}{x^2} \rightarrow 0 \text { as } x \rightarrow \infty \\ & \Rightarrow \lim _{x \rightarrow \infty} \frac{5}{1-\frac{2}{x^2}}-\lim _{x \rightarrow \infty} \frac{\frac{\sin x}{x^2}}{1-\frac{2}{x^2}} \\ & \qquad \sin x=\frac{1}{\operatorname{cosec} x} \\ & \Rightarrow \lim _{x \rightarrow \infty} \frac{5}{1-2 / x^2}-\lim _{x \rightarrow \infty} \frac{\frac{1}{x^2 \operatorname{cosec} x}}{1-2 / x^2} \\ & \Rightarrow 5-0=5 \end{aligned} $

We have,

$ \lim \limits_{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\frac{0}{0}) $

Using L,Hospital's rule,

$ \begin{aligned} & \lim _{x \rightarrow 2} \frac{\frac{1}{2} \frac{4}{\sqrt{1+4 x}}-\frac{1}{2} \frac{3}{\sqrt{3+3 x}}}{3 x^2} \\ & =\lim _{x \rightarrow 2} \frac{4 \sqrt{3+3 x}-3 \sqrt{1+4 x}}{6 x^2 \sqrt{3+3 x} \sqrt{1+4 x}} \\ & =\frac{4 \times 3-3 \times 3}{6 \times 4 \times 3 \times 3}=\frac{3}{72 \times 3}=\frac{1}{72} \end{aligned} $

Therefore, the limit is $1 / 72$.

$ \begin{aligned} & \quad[\sqrt{2 x+1}+\sqrt{2 x-1})^8 \\ & \lim _{x \rightarrow \infty} \frac{\left.+(\sqrt{2 x+1}-\sqrt{2 x-1})^8\right]\left(p x^4-18\right.}{\left(x+\sqrt{x^2-2}\right)^8+\left(x-\sqrt{x^2-2}\right)^8} =1 \end{aligned} $

Take the ' $x$ ' from numerator and denominator.

$ (x)^4\left(\sqrt{2+\frac{1}{x}}+\sqrt{2-\frac{1}{x}}\right)^8 $

$ \begin{aligned} & \lim _{x \rightarrow \infty} \frac{+\left(\sqrt{2+\frac{1}{x}}-\sqrt{2-\frac{1}{x}}\right)^8 \cdot x^4\left(p-\frac{16}{x}\right)}{x^8\left[\left(1+\sqrt{2+\frac{1}{x}}\right)^8+\left(1-\sqrt{2-\frac{1}{x}}\right)\right.} =1 \end{aligned} $

Canceled $x^8$ from numerator and denominator and put the value of $ x \rightarrow \infty $

$ \begin{aligned} & \frac{\left\{[\sqrt{(2+0)}+\sqrt{2-0}]^8+[\sqrt{2}-\sqrt{2}]^8\right\}\{P-0}{(1+\sqrt{1})^8+(1-\sqrt{1})^8} =1 \\ & \frac{\left[(\sqrt{2}+\sqrt{2})^8+(0)^8\right]\{P\}}{(2)^8+(0)^8}=1 \\ & \frac{(2 \sqrt{2})^8 P}{(2)^8}=1 \end{aligned} $

$ \begin{aligned} P=\frac{(2)^8}{(2 \sqrt{2})^8}=\frac{2^8}{2^8 \cdot(\sqrt{2})^8} \\ P=\frac{1}{2^4}=\frac{1}{16} \Rightarrow P=\frac{1}{16} \end{aligned} $

$ \begin{aligned} \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x} \end{aligned} $

On putting the limit $x \rightarrow \frac{\pi}{4}$, we get $\frac{0}{0}$

which is an indeterminate form, using

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

$\lim \limits_{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$

$ \text { LHS }=\lim \limits_{x \rightarrow 0} \frac{\frac{\left\{e^x-a-\log (1+x)\right\}}{x}}{\frac{\sin x}{x}} $

Dividing by $x$ to both numerator and denominator

$\begin{aligned} & =\frac{\lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}-\frac{\log (1+x)}{x}}{\lim \limits_{x \rightarrow 0} \frac{\sin x}{x}} \\ & =\frac{\lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}-\lim \limits_{x \rightarrow 0} \frac{\log (1+x)}{x}}{\lim \limits_{x \rightarrow 0} \frac{\sin x}{x}} \\ & =\frac{\lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}-1}{1}\end{aligned}$

$ \left[\because \lim \limits_{x \rightarrow 0} \frac{\log (1+x)}{x}=1, \lim \limits_{x \rightarrow 0} \frac{\sin x}{x}=1\right] $

$ =0=\text { RHS } $

$ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{e^x-a}{x}-1=0 $

$ \Rightarrow \quad \lim \limits_{x \rightarrow 0} \frac{e^x-a}{x}=1 $

$ \therefore \quad a=1 \quad\left(\because \lim \limits_{x \rightarrow 0} \frac{e^x-1}{x}=1\right) $

Given,

$ f(x)=\left\{\begin{array}{cc}(1+|\sin x|)^{\frac{a}{|\sin x|}}, & -\frac{\pi}{6} < x < 0 \\ b, & x=0 \\ e^{\frac{\tan 2 x}{\tan 3 x}}, & 0 < x < \frac{\pi}{6}\end{array}\right. $

is continuous at $x=0$

$ \lim \limits_{x \rightarrow 0^{-}} f(x)=f(0)=\lim \limits_{x \rightarrow 0^{+}} f(x) $

Now,

$ \begin{gathered} \begin{aligned} \lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{\frac{a}{|\sin x|}} & =e^{\lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}} \\ & =e^a \\ \lim _{x \rightarrow 0^{+}} e^{\frac{\tan 2 x}{\tan 3 x}} & =\lim _{x \rightarrow 0} e^{\left(\frac{\tan 2 x}{2 x} \times \frac{3 x}{\tan 3 x} \times \frac{2 x}{3 x}\right)} \\ & =\lim _{x \rightarrow 0} e^{2 / 3}=e^{2 / 3} \end{aligned} \end{gathered} $

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

$ e^a=b=e^{2 / 3} \Rightarrow a=\frac{2}{3} $

and $b=e^{2 / 3}$

$f(x)=\left\{\begin{aligned} 2 x+3, & x \leq 1 \\ a x^2+b x, & x>1\end{aligned}\right.$

Given $f$ is differentiable for all $x \in R$.

$\therefore f$ is continuous everywhere.

At $x=1$

LHL $=$ RHL

$\lim \limits_{x \rightarrow 1^{-}} f(x)=\lim \limits_{x \rightarrow 1^{+}} f(x)$

$\Rightarrow \quad \lim \limits_{x \rightarrow 1^{-}} 2 x+3=\lim \limits_{x \rightarrow 1^{+}} a x^2+b x$

$\Rightarrow \quad 2(1)+3=a(1)^2+b(1)$

$\Rightarrow \quad 5=a+b\quad \ldots \text { (i) }$

Also, LHD = RHD

$ \begin{array}{ll} & \lim \limits_{x \rightarrow 1^{-}} f^{\prime}(x)=\lim \limits_{x \rightarrow 1^{+}} f^{\prime}(x) \\ \Rightarrow \quad & 2=\lim \limits_{x \rightarrow 1^{+}} 2 a x+b \\ \Rightarrow \quad & 2=2 a+b \quad \ldots \text { (i) }\end{array} $

From Eqs. (i) and (ii), we get

$\begin{array}{ll} & a=-3, b=8 \\ \therefore & f(x)=\left\{\begin{array}{cc}2 x+3, & x \leq 1 \\ -3 x^2+8 x, & x>1\end{array}\right.\end{array}$

Now, $f^{\prime}(x)=\left\{\begin{array}{cc}2, & x \leq 1 \\ -6 x+8, & x>1\end{array}\right.$

So, $\quad f^{\prime}(2)=-6(2)+8=-12+8=-4$

Given $f(x)=|x-1|+|x-2|$

$ \begin{aligned} f(x) & = \begin{cases}(1-x)+(2-x), & 0 \leq x<1 \\ (x-1)+(2-x), & 1 \leq x<2 \\ (x-1)+(x-2), & 2 \leq x \leq 3\end{cases} \\ & =\left\{\begin{array}{cc} 3-2 x, & 0 \leq x<1 \\ 1, & 1 \leq x<2 \\ 2 x-3, & 2 \leq x \leq 3 \end{array}\right. \end{aligned} $

At $x=1$

$ \begin{aligned} & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3-2 x=3-2(1)=1 \\ & \qquad f(1)=1, \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 1=1 \\ & \therefore \lim _{x \rightarrow 1^{-}} f(x)=f(1)=\lim _{x \rightarrow 1^{+}} f(x) \end{aligned} $

Now, $\lim _{x \rightarrow 1^{-}} f^{\prime}(x)=\lim _{x \rightarrow 1^{-}}(3-2 x)^{\prime}$

$ \begin{gathered} =\lim _{x \rightarrow 1^{-}}-2=-2 \\ \lim _{x \rightarrow 1^{+}} f^{\prime}(x)=0 \\ \therefore \lim _{x \rightarrow 1^{-}} f^{\prime}(x) \neq \lim _{x \rightarrow 1^{+}} f^{\prime}(x) \end{gathered} $

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

$ \text { At, } x=2 $

$ \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 1=1 $

$ f(2)=2(2)-3=1 $

$ \lim \limits_{x \rightarrow 2^{+}} f(x)=\lim \limits_{x \rightarrow 2^{+}}(2 x-3)=4-3=1 $

$ \therefore \lim \limits_{x \rightarrow 2^{-}} f(x)=f(2)=\lim \limits_{x \rightarrow 2^{+}} f(x) $

$ \lim \limits_{x \rightarrow 2^{-}} f^{\prime}(x)=0, \lim \limits_{x \rightarrow 2^{+}} f^{\prime}(x)=\lim \limits_{x \rightarrow 2^{+}} 2 $

$ \therefore \lim \limits_{x \rightarrow 2^{-}} f^{\prime}(x) \neq \lim \limits_{x \rightarrow 2^{+}} f^{\prime}(x) $

So, $f(x)$ is continuous but not differentiable at $x=2$

$ \begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4} \\ & =\lim _{x \rightarrow 0} \frac{\left[\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}\right]\left[\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}\right]}{x^4\left[\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}\right]} \\ & =\lim _{x \rightarrow 0} \frac{1+\sqrt{1+x^4}-2-x^5-x^6}{x^4} \times-\lim _{x \rightarrow 0} \frac{1}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}} \\ & =\lim _{x \rightarrow 0}\left[\frac{\sqrt{1+x^4}-1}{x^4}-\frac{x^5(1+x)}{x^4}\right] \times \frac{1}{2 \sqrt{x}} \\ & =\frac{1}{2 \sqrt{2}} \lim _{x \rightarrow 0} \frac{\left(\sqrt{1+x^4}-1\right)\left(\sqrt{1+x^4}+1\right)}{x^4\left(\sqrt{1+x^4}+1\right)}=\frac{1}{2 \sqrt{2}} \times \frac{1}{2}=\frac{1}{4 \sqrt{2}} \end{aligned} $

$ =\lim \limits_{x \rightarrow 1} \frac{\frac{1}{2 \sqrt{x}}}{2 \cos ^{-1} x \times\left(\frac{-1}{\sqrt{1-x^2}}\right)}=\lim \limits_{x \rightarrow 1} \frac{-\sqrt{1-x^2}}{4 \sqrt{x} \cos ^{-1} x}\left(\frac{0}{0}\right) $

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

To ensure the function $ f(x) $ is continuous at $ x = 0 $, the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at zero must all be equal:

$ \text{LHL} = \text{RHL} = f(0) $

Calculating the Left-Hand Limit (LHL):

For $ x < 0 $, we have:

$ f(x) = \frac{\sin 3x - \tan 3x}{x^3} $

The LHL is evaluated as follows:

$ \begin{aligned} \lim_{x \to 0^{-}} \frac{\sin 3x - \tan 3x}{x^3} &= \lim_{x \to 0^{-}} \frac{\sin 3x(\cos 3x - 1)}{\cos 3x \times x^3} \\ &= \lim_{x \to 0^{-}} \frac{\sin 3x}{x} \times \lim_{x \to 0^{-}} \frac{1}{\cos 3x} \times \lim_{x \to 0^{-}} \frac{\cos 3x - 1}{x^2} \\ &= 3 \times 1 \times \lim_{x \to 0^{-}} (-2) \frac{\sin^2 \frac{3x}{2}}{\frac{9x^2}{4}} \times \frac{9}{4} \\ &= 3 \times (-2) \times \frac{9}{4} = -\frac{27}{2} \end{aligned} $

Calculating the Right-Hand Limit (RHL):

For $ x > 0 $, the function is:

$ f(x) = \frac{\tan((\alpha + 1)x) + \tan(2x)}{x} $

The RHL is computed as:

$ \begin{aligned} \lim_{x \to 0^{+}} \frac{\tan((\alpha + 1)x) + \tan(2x)}{x} &= \lim_{x \to 0^{+}} \frac{\tan((\alpha + 1)x)}{x} + \lim_{x \to 0^{+}} \frac{\tan(2x)}{x} \\ &= \alpha + 1 + 2 = \alpha + 3 \end{aligned} $

Value at $ x = 0 $:

$ f(0) = \beta $

Equating Limits for Continuity:

Given that LHL = RHL = $ f(0) $, we have:

$ \begin{aligned} -\frac{27}{2} &= \alpha + 3 = \beta \\ \alpha &= -\frac{27}{2} - 3 = -\frac{33}{2} \\ \beta &= -\frac{27}{2} \end{aligned} $

Calculate $ |\alpha| + |\beta| $:

$ |\alpha| = \frac{33}{2}, \quad |\beta| = \frac{27}{2} $

$ |\alpha| + |\beta| = \frac{33}{2} + \frac{27}{2} = \frac{60}{2} = 30 $

$ \begin{aligned} &\text {We have, } \lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9}\\ &\begin{aligned} & =\lim _{x \rightarrow 3} \frac{(x-3)\left(x^2+9+3 x\right)}{(x-3)(x+3)} \\ & =\lim _{x \rightarrow 3} \frac{x^2+9+3 x}{x+3} \\ & =\frac{9+9+9}{6}=\frac{27}{6}=\frac{9}{2} \end{aligned} \end{aligned} $

Given the function $ f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases} $, we need to determine the condition for the existence of the limit $ \lim \limits_{x \rightarrow 1} f(x) $.

For the limit to exist at $ x = 1 $, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal:

$ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) $

Calculating the limits from both sides:

Left-hand limit (x approaches 1 from the left):

$ LHL = a(1) + b + 1 = a + b + 1 $

Right-hand limit (x approaches 1 from the right):

$ RHL = 3a(1) - 2b = 3a - 2b $

Setting the LHL equal to the RHL:

$ a + b + 1 = 3a - 2b $

Solving for the relationship between $ a $ and $ b $:

$ a + b + 1 = 3a - 2b $

Rearranging terms yields:

$ 2a - 3b = 1 $

Thus, the relation between $ a $ and $ b $ when the limit exists is $ 2a - 3b = 1 $.

We have, $f(x)= \begin{cases}\frac{2}{5-x}, & x<3 \\ 5-x, & x \geq 3\end{cases}$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}} \frac{2}{5-x}=1 \\ & \mathrm{RHL}=\lim _{x \rightarrow 3^{+}} f(x)=5-3=2 \end{aligned} $

Hence, the function is left discontinuous at $x=3$.

We have, $f(x)=\left\{\begin{array}{cc}x^\alpha \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.$

$ \begin{aligned} & \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{x^a\left(\sin \frac{1}{x}\right)}{\left(\frac{1}{x}\right)} \times \frac{1}{x} \\ & =\lim _{x \rightarrow 0} x^{\alpha-1}\left[\because \lim _{x \rightarrow 0} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}=1\right]=0 \end{aligned} $

$\therefore f(x)$ is continuous for all value except -1 .

Now, $f(x)$ is differentiable at $x=0$, if $\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$ exists.

$\Rightarrow \quad \lim _{x \rightarrow 0} x^{\alpha-1} \sin \frac{1}{x}$ exists finitely.

$ \Rightarrow \quad \alpha-1>0 $

$ \Rightarrow \quad \alpha>1 $

Hence, $f(x)$ is continuous and differentiable for $\alpha>1$.

The graph of $f(x)=\min \left\{x, x^2\right\}$ is shown alongside.

From the diagram, it is clear that $f(x)$ is continuous everywhere but $f(x)$ is not differentiable but $x=0$ and $x=1$.

$ f(x)= \begin{cases}x, & \text { if } x<0 \\ x^2, & \text { if } 01\end{cases} $

For $x>1, f^{\prime}(x)=1$

Given,

$ \begin{gathered} f(x)=\left\{\begin{array}{l} \frac{2 x e^{\frac{1}{2 x}}-3 x e^{-\frac{1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{-\frac{1}{2 x}}} \text { If } x \neq 0 \\ 0 \quad \text { If } x=0 \end{array}\right. \\ \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 x e^{1 / 2 x}-3 x e^{-1 / 2 x}}{e^{1 / 2 x}+4 e^{-1 / 2 x}} \\ =\lim _{x \rightarrow 0} \frac{x\left(2 e^{1 / x}-3\right)}{e^{1 / x}+4}=0 \end{gathered} $

$\therefore f(x)$ is continuous of $x=0$

$ \begin{aligned} & \text { LHD } \begin{aligned} f^{\prime}\left(0^{-}\right) & =\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{h\left(-2 e^{-1 / h}-3\right)}{-h\left(e^{-1 / h}+4\right)} \\ & =\lim _{h \rightarrow 0} \frac{2 e^{-1 / h}-3}{e^{-1 / h}+4}=\frac{-3}{4} \end{aligned} \end{aligned} $

$\begin{aligned} & f\left(0^{-}\right)=\frac{-3}{4} \\ & \begin{aligned} \text { RHD }=f^{\prime}\left(0^{+}\right) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{h\left(2 e^{1 / h}-3\right)}{h\left(e^{1 / h}+4\right)} \\ & =\lim _{h \rightarrow 0} \frac{e^{1 / h}\left(2-\frac{3}{e^{1 / h}}\right)}{e^{1 / h}\left(1+\frac{4}{e^{1 / h}}\right)}=2\end{aligned} \\ & \text { RHD }=f^{\prime}\left(0^{-}\right)=2\end{aligned}$ LHD $\neq$ RHD $\quad \begin{aligned} & \therefore f \text { is not different of } x=0\end{aligned}$

The Taylor series for $e^x$, $\cos x$, and $e^{-x}$ around $x=0$ are:

$e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,$

$\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,$

$e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.$

Substituting these into the limit and simplifying:

$\lim\limits_{x \rightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}.$

This simplifies to:

$\lim\limits _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}.$

Let's evaluate this expression:

$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2.$

We know that $1 - \cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\cos 2x = 1 - 2\sin^2x$. So, $1 - \cos 2x = 2\sin^2x$.

Substituting this into our expression we get:

$(\frac{2\sin^2x}{(2x)^2}) \times 4x^2.$

This simplifies to:

${1 \over 2}$$(\frac{\sin x}{x})^2 \times 4x^2 = {1 \over 2} \times 4x^2 = 2x^2.$

So,

$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2 = 2x^2.$

$ \therefore $ $\lim _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17.$

Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:

$a-\frac{c}{2} = 0 \Rightarrow c=2a.$

Then, for the limit to equal $17$, the coefficient of $x^2$ in the numerator must equal $17 \times 2 = 34$. This gives us the equation:

$\frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68.$

Since $c=2a$, we can substitute $c$ in the equation to get:

$a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.$

$f(x) = {\mathop{\rm sgn}} (x)$

$h(x) = 2[x] - {\mathop{\rm sgn}} (x)$

If $x \to {1^ + }$ then $h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$

$ = 0 - 1 = - 1$

& if $x \to {1^ - }$ then $h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$

$ = - 2 + 1 = - 1$

$\therefore$ $\mathop {\lim }\limits_{x \to {1^ + }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ + }} {{\sin (h(x + 1) + 1)} \over {h(x - 1) + 1}} = 1$

$\mathop {\lim }\limits_{x \to {1^ - }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ - }} {{\sin (h(x - 1) + 1)} \over {h(x - 1) + 1}} = 1$

$5f(x + y) = f(x).f(y)$

$5f(3) = f(1).f(2)$

$5f(2) = {(f(1))^2}$

$f(10) = 5$

$f(1) = 20$

$ \Rightarrow f(1).{{{{(f(1))}^2}} \over 5} = 1600$

$\sum\limits_{n = 0}^5 {f(n) = f(0) + 20 + 80 + 320 + 1280 + 5120} $

$ = 1750 + 5120 = 6825$

$\mathop {\lim }\limits_{x \to a} \left( {[x - 5] - [2x + 2]} \right) = 0$

$ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {[x] - 5 - [2x] - 2} \right) = 0$

$ \Rightarrow \mathop {\lim }\limits_{x \to a} [x] - [2x] - 7 = 0$

Case 1 :

If $a \in \left[ {n,n + {1 \over 2}} \right)$

then $\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0$

$ \Rightarrow n - 2n - 7 = 0$

$ \Rightarrow n = - 7$

$\therefore$ $a \in \left[ { - 7, - 6.5} \right)$

Case 2 :

If $a \in \left[ {n + {1 \over 2},n + 1} \right)$

then $\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0$

$ \Rightarrow n - (2n + 1) - 7 = 0$

$ \Rightarrow - n - 8 = 0$

$ \Rightarrow n = - 8$

$\therefore$ $a \in \left[ { - 7.5, - 7} \right)$

$R.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ + }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right)$

$ = - 8 - \left( { - 15} \right) - 7$

$ = - 8 + 15 - 7 = 0$

$\eqalign{ & L.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ - }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right) \cr \\ & = - 8 - \left( { - 16} \right) - 7 \cr \\ & = - 8 + 16 - 7 = 1 \cr} $

$ \therefore $ R.H.L $ \ne $ L.H.L

$ \therefore $ At x = -7.5, limit does not exists.

$\therefore$ $a \in \left( { - 7.5, - 6.5} \right)$

Given,

$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right),} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$

$\therefore$ $f'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$

Now,

$\mathop {\lim }\limits_{x \to 0} f'(x) = \mathop {\lim }\limits_{x \to 0} \left[ {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right]$

$ = 0 - \mathop {\lim }\limits_{x \to 0} \cos \left( {{1 \over x}} \right)$

= Does not exist

$\therefore$ $f'(x)$ is discontinuous function at $x = 0$.

L.H.D. $ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 - h) - f(0)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f( - h)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} {{ - {h^2}\sin \left( {{1 \over h}} \right)} \over { - h}} = 0$

R.H.D. $ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 + h) - f(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(h)} \over h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} {{{h^2}\sin \left( {{1 \over h}} \right)} \over h} = 0$

$\therefore$ L.H.D. = R.H.D.

$ \Rightarrow f(x)$ is differentiable at $x = 0$. So, f(x) is continuous.