Limits, Continuity and Differentiability

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 { } \because \lim _{x \rightarrow 1} f(x) \text { exists } \\ & \therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \end{aligned} $

$ \begin{aligned} &Now, \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ & =\lim _{h \rightarrow 0}\left(1+\frac{2(1-h)}{a}\right)=1+\frac{2}{a} \end{aligned} $

$ \begin{array}{ll} \text { and } & \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ & =\lim _{h \rightarrow 0} a(1+h)=a \\ \therefore & 1+\frac{2}{a}=a \\ \Rightarrow & a+2=a^2 \\ \Rightarrow & a^2-a-2=0 \\ \Rightarrow & a^2-2 a+a-2=0 \\ \Rightarrow & a(a-2)+1(a-2)=0 \\ \Rightarrow & (a-2)(a+1)=0 \\ \Rightarrow & a=-1,2 \end{array} $

$\therefore$ Sum of the cubes of the possible values of $a$ is $-1+8=7$.

At $x=0$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}}\left(\left[x^2\right]-[x]^2\right)=0-1=-1 \\ & \mathrm{RHL}=\lim _{x \rightarrow 0^{+}}\left(\left[x^2\right]-[x]^2\right)=0-0=0 \end{aligned} $

$\therefore$ LHL $\neq$ RHL

Limit does not exists at $x=0$.

At $x=1$

$ \mathrm{LHL}=\lim _{x \rightarrow 1^{-}}\left(\left[x^2\right]-[x]^2\right)=0-0=0 $

$\mathrm{RHL}=\lim _{x \rightarrow 1^{+}}\left(\left[x^2\right]-[x]^2\right)=1-1=0$

As LHL $=$ RHL, so limit exits at $x=1$

At $x=2$

$ \begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 2^{-}}\left(\left[x^2\right]-[x]^2\right)=3-1=2 \\ & \mathrm{RHL}=\lim _{x \rightarrow 2^{+}}\left(\left[x^2\right]-[x]^2\right)=4-4=0 \end{aligned} $

As LHL $\neq$ RHL, so limit does not exists. Hence, there are two integral points where the limits does not exists.

$ \begin{aligned} &\text { As } f(x) \text { is continuous at } x=0\\ &\begin{aligned} & f(0)=\lim _{x \rightarrow 0} f(x) \\ & K=\lim _{x \rightarrow 0} \frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ & \Rightarrow K=\lim _{x \rightarrow 0} \\ & \frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ & \quad \quad \times \frac{\sqrt{a^2-a x+x^2}+\sqrt{x^2+a x+a^2}}{\left.\sqrt{a^2-a x+x^2}+\sqrt{x^2+a x+a^2}\right)} \\ & \quad \quad \times \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-2 a x(\sqrt{a+x}+\sqrt{a-x)}}{2 x\left(\sqrt{x^2-a x+a^2}+\sqrt{x^2+a x+a^2}\right)} \\ & =\frac{-a(2 \sqrt{a})}{2 a}=-\sqrt{a} \\ & \therefore \quad K=-\sqrt{a} \end{aligned} $

$f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 < x \leq 2 \\ b x^3+1, & x > 2\end{array}\right.$

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

As, the function is differentiable at $x \in R$,

$ \begin{aligned} & \lim _{x \rightarrow 1} 2 a x+b=3 \text { and } \lim _{x \rightarrow 2} 3 b x^2=3 \\ & \Rightarrow 2 a+b=3 \text { and } 12 b=3 \Rightarrow b=\frac{1}{4} \\ & \quad 2 a=3-\frac{1}{4} \Rightarrow 2 a=\frac{11}{4} \Rightarrow a=\frac{11}{8} \\ & \therefore \quad a-b=\frac{11}{8}-\frac{1}{4}=\frac{9}{8} \end{aligned} $

For option (a), $f(x)=|x|=x$ as $1 \leq x \leq 5$

As $f(x)$ is a polynomial function in its domain, so it is continuous and differentiable.

$ \begin{aligned} & f^{\prime}(x)=1, f(1)=1 \text { and } f(3)=5 \\ & f^{\prime}(x)=\frac{f(5)-f(1)}{5-1}=1 \in[1,5] \end{aligned} $

$\therefore$ Lagrange's mean value theorem is applicable.

For option (b)

$ f(x)=[x]=1 \text { for } x \in[\sqrt{2}, \sqrt{3}] $

As $f(x)$ is a constant function, so Lagranye's mean value theorem is applicable.

For option (c),

The function $f(x)=x^2-1$ is continuous if $x^2-1>0$ i.e. either $x<-1$ or $x>1$.

For $x=\frac{1}{e}, x^2-1>0$, but $\frac{1}{e^2}-1<0$

So, the value is not within the domain Thus, lagrange mean value theorem is not applicable.

For option (d), $f(x)=e^x$ is continuous and differentiable over $[-e, e]$

$ f^{\prime}(x)=e^x \text { and } f(-e)=\frac{1}{e^i} f(l)=e^e{} $

$ \begin{aligned} f^{\prime}(e) & =\frac{f(b)-f(a)}{b-a}=\frac{e^e-\frac{1}{e^e}}{e+e} \\ & =\frac{\left(e^e\right)^2-1}{e^e \times 2 e} \\ & =\frac{e^{e-1}}{2}-\frac{1}{2 e^{e+1}} \in[-e, e] \end{aligned} $

$\therefore$ Lagrange mean value theorem is applicable.

$ \begin{aligned} & f(x)=\left\{\begin{array}{cc} \frac{x-|x|}{x}, & x \neq 0 \\ 2, & x=0 \end{array}\right. \\ & = \begin{cases}2, & x<0 \\ 0, & x>0 \\ 2, & x=0\end{cases} \\ & \quad\left[\because|x|=\left\{\begin{array}{c} x, \text { if } x>0 \\ -x, \text { if } x<1 \end{array}\right.\right. \end{aligned} $

$\therefore f(x)$ has maximum value 2 and minimum value 0 .

$\mathop {\lim }\limits_{x \to \infty } \frac{[2 x-3]}{x} \text { } $

$ \begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{2 x}{x}-\frac{3}{x}\right) \Rightarrow 2-\lim _{x \rightarrow \infty} \frac{3}{x} \\ & \Rightarrow \quad 2-0 \Rightarrow 2 \end{aligned} $

$\mathop {\lim }\limits_{x \to 0} \frac{\cos 2 x-\cos 3 x}{\cos 4 x-\cos 5 x} \quad\left(\frac{0}{0}\right.$ form $)$

By L' Hospital rule,

$ =\mathop {\lim }\limits_{x \to 0} \frac{-2 \sin 2 x+3 \sin 3 x}{-4 \sin 4 x+5 \sin 5 x} \quad\left(\frac{0}{0} \text { form }\right) $

Again by L' Hospital rule

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-4 \cos 2 x+9 \cos 3 x}{-16 \cos 4 x+25 \cos 5 x} \\ & =\frac{-4 \times 1+9 \times 1}{-16 \times 1+25 \times 1}=\frac{5}{9} \end{aligned} $

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 $\quad \begin{aligned} & \therefore f \text { is not different of } x=0\end{aligned}$

$ f(x)=\left\{\begin{array}{cc}\frac{x^{2}-1}{x^{2}-2 x+1} ; & x > 1 \\ \frac{1}{2} ; & x=1 \\ \frac{x^{2}-1}{x^{2}+2 x-3} ; & x < 1\end{array}\right. $

$ \begin{array}{ll} \because & x^{2}+3 x-x-3 \\ & =x(x+3)-1(x+3) \end{array} $

Lets Find

LHL, $ \lim \limits_{x \rightarrow 1^{-}} \frac{x^{2}-1}{x^{2}-2 x+1} $ and RHL, $ \lim \limits_{x \rightarrow 1^{+}} \frac{x^{2}-1}{x^{2}+2 x-3} $

$ \begin{array}{l} \Rightarrow \lim \limits_{x \rightarrow 1^{-}} \frac{(x-1)(x+1)}{(x-1)^{2}} \Rightarrow \lim \limits_{x \rightarrow 1^{+}} \frac{(x-1)(x+1)}{(x+3)(x-1)} \\ \Rightarrow \lim \limits_{x \rightarrow 1^{-}} \frac{x+1}{x-1}=\infty \Rightarrow \lim \limits_{x \rightarrow 1^{+}} \frac{x+1}{x+3}=\frac{2}{4}=\frac{1}{2} \end{array} $

$ \therefore \mathrm{LHL} \neq f(1) \Rightarrow f(x) $ is not continuous at $ x=1 $

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:

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