Limits, Continuity and Differentiability

The function is defined as :

$ f(x)=\lim\limits_{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right) $

Let $n=\frac{2}{\theta}$. As $\theta \rightarrow 0, n \rightarrow \infty$ (specifically, let's consider $\theta$ approaching 0 such that the power grows). The behavior of the function depends on the value of $x$ because of the term $x^n$.

• Case 1 : $|x|<1$ As $n \rightarrow \infty, x^n \rightarrow 0$.

$ f(x)=\frac{\cos \pi x-0}{1+0}=\cos \pi x $

• Case 2 : $|x|>1$ As $n \rightarrow \infty, x^n \rightarrow \infty$. We divide the numerator and denominator by $x^n$ :

$ f(x)=\lim\limits_{n \rightarrow \infty} \frac{\frac{\cos \pi x}{x^n}-\sin (x-1)}{\frac{1}{x^n}+(x-1)}=\frac{0-\sin (x-1)}{0+(x-1)}=\frac{-\sin (x-1)}{x-1} $

• Case 3 : $x=1$ If $x=1$, then $\cos (\pi)=-1$ and $\sin (0)=0$.

$ f(1)=\frac{-1-\left(1^n \cdot 0\right)}{1+\left(1^n \cdot 0\right)}=-1 $

• Case 4 : $x=-1$ If $x=-1$, then $\cos (-\pi)=-1$ and $\sin (-2)=-\sin (2)$. Since $n= 2 / \theta$ is even, $(-1)^n=1$.

$ f(-1)=\frac{-1-(1 \cdot \sin (-2))}{1+(1 \cdot-2)}=\frac{-1+\sin 2}{-1}=1-\sin 2 $

So $\mathrm{f}(\mathrm{x})$ become

$ f(x)= \begin{cases}\frac{-\sin (x-1)}{x-1} & \text { if } x \in(-\infty,-1) \cup(1, \infty) \\ \cos \pi x & \text { if } x \in(-1,1) \\ 0 & \text { if } x=1 \\ 1-\sin 2 & \text { if } x=-1\end{cases} $

Analyze the Statements

Statement (I) : $f(x)$ is discontinuous at $x=1$.

• Value at $x=1$ : $f(1)=-1$.

• Left-hand limit ( $x \rightarrow 1^{-}$) : Uses the $|x|<1$ definition: $\lim _{x \rightarrow 1^{-}} \cos \pi x=\cos \pi=-1$.

• Right-hand limit $\left(x \rightarrow 1^{+}\right)$ : Uses the $|x|>1$ definition: $\lim\limits_{x \rightarrow 1^{+}} \frac{-\sin (x-1)}{x-1}$. Using the standard limit $\lim\limits_{u \rightarrow 0} \frac{\sin u}{u}=1$, this equals -1. Since $L H L=R H L=f(1)$, the function is continuous at $x=1$.
  
  Conclusion : Statement (I) is False.

Statement (II) : $f(x)$ is continuous at $x=-1$.

• Value at $x=-1$ : $f(-1)=1-\sin 2$.

• Left-hand limit $\left(x \rightarrow-1^{-}\right)$: Uses $|x|>1$ definition : $\frac{-\sin (-1-1)}{-1-1}=\frac{-\sin (-2)}{-2}=\frac{\sin 2}{-2}$.

• Right-hand limit ( $x \rightarrow-1^{+}$): Uses $|x|<1$ definition : $\cos (-\pi)=-1$.
  
  Since the limits do not match, the function is discontinuous at $x=-1$.
  
  Conclusion : Statement (II) is False.

Let $L=\lim\limits_{x \rightarrow 0} \frac{\ln \left(\sec (e x) \cdot \sec \left(e^2 x\right) \cdots \cdots \sec \left(e^{10} x\right)\right)}{e^2-e^{2 \cos x}}$

Simplify : we use $\ln (a b c \ldots)=\ln (a)+\ln (b)+\ln (c) \ldots$ and take $e^2$ common in denominator

$ L=\lim\limits_{x \rightarrow 0} \frac{\ln \sec (e x)+\ln \sec \left(e^2 x\right)+\cdots+\ln \sec \left(e^{10} x\right)}{e^2\left[1-e^{2 \cos x-2}\right]} $ ........(1)

Now understanding dominating term

• If $A \rightarrow 1 \Rightarrow \ln A \approx A-1$

• as $x \rightarrow 0 \Rightarrow \sec (e x) \rightarrow 1 \Rightarrow \ln \sec (e x) \approx \sec (e x)-1$

• so generalized $x \rightarrow 0 \Rightarrow \ln \sec \left(e^k x\right) \approx \sec \left(e^k x\right)-1$

• If $A \rightarrow 0 \Rightarrow e^A-1 \approx A$

• so in denominator as $x \rightarrow 0 \Rightarrow 2(\cos x-1)=0$

• so, $e^{2(\cos x-1)}-1 \approx 2(\cos x-1)$

Substituting this dominating term in equation (1)

$ \begin{aligned} & L=\lim\limits_{x \rightarrow 0} \frac{(\sec (e x)-1)+\left(\sec \left(e^2 x\right)-1\right)+\cdots+\left(\sec \left(e^{10} x\right)-1\right)}{-e^2 \cdot 2(\cos x-1)} \\ = & \lim _{x \rightarrow 0} \frac{\left(\frac{1}{\cos (e x)}-1\right)+\left(\frac{1}{\cos \left(e^2 x\right)}-1\right)+\cdots+\left(\frac{1}{\cos \left(e^{10} x\right)}-1\right)}{2 e^2(1-\cos x)}=\lim _{x \rightarrow 0} \frac{\frac{1-\cos (e x)}{\cos (e x)}+\frac{1-\cos \left(e^2 x\right)}{\cos \left(e^2 x\right)}+\cdots+\frac{1-\cos \left(e^{10} x\right)}{\cos \left(e^{10} x\right)}}{2 e^2(1-\cos x)} \\ = & \lim _{x \rightarrow 0} \frac{1-\cos (e x)}{\cos (e x) \cdot 2 e^2(1-\cos x)}+\lim _{x \rightarrow 0} \frac{1-\cos \left(e^2 x\right)}{\cos \left(e^2 x\right) \cdot 2 e^2(1-\cos x)}+\cdots+\frac{1-\cos \left(e^{10} x\right)}{\cos \left(e^{10} x\right) \cdot 2 e^2(1-\cos x)} ........(2) \end{aligned} $

Again using dominating term

at $x \rightarrow 0, \cos \left(e^k x\right)=1, k=\{1,2, \ldots, 10\}$

$ \begin{aligned} L & =\lim _{x \rightarrow 0} \frac{1-\cos (e x)}{2 e^2(1-\cos x)}+\lim _{x \rightarrow 0} \frac{1-\cos \left(e^2 x\right)}{2 e^2(1-\cos x)}+\cdots+\frac{1-\cos \left(e^{10} x\right)}{2 e^2(1-\cos x)} \end{aligned} $

Understanding dominating term again

If $A \rightarrow 0 \Rightarrow 1-\cos A \approx \frac{A^2}{2}$ So $1-\cos \left(e^k x\right) \approx \frac{\left(e^k x\right)^2}{2}, k \in\{1,2, \ldots, 10\}$

Substitute this value in equation (2) :

$L=\frac{\lim\limits_{x \rightarrow 0} \frac{(e x)^2}{2}}{2 e^{2}{\cdot \frac{x^2}{2}}}+\frac{\lim\limits_{x \rightarrow 0} \frac{\left(e^2 x\right)^2}{2}}{2 e^{2}{ \cdot \frac{x^2}{2}}}+\cdots+\frac{\lim\limits_{x \rightarrow 0} \frac{\left(e^{10} x\right)^2}{2}}{2 e^{2}{ \cdot \frac{x^2}{2}}}$

$x^2$ and $\frac{1}{2}$ get cancelled so $L$ becomes :

$L=\frac{e^2}{2 e^2}+\frac{\left(e^2\right)^2}{2 e^2}+\cdots+\frac{\left(e^{10}\right)^2}{2 e^2}$

take $\frac{1}{2 e^2}$ common $L=\frac{1}{2 e^2}\left[e^2+e^4+e^6+\cdots+e^{20}\right]$

$e^2+e^4+\cdots+e^{20}$ is a G.P. whose first term is $e^2$ common ratio is $e^2$ so sum of G.P. is $a \frac{\left(r^n-1\right)}{r-1}$ where $a$ is first term and $r$ is common ratio.

$L=\frac{1}{2 e^2} \times e^2\left[\frac{\left(e^2\right)^{10}-1}{e^2-1}\right] L=\frac{e^{20}-1}{2\left(e^2-1\right)}$

$ \lim\limits_{t \rightarrow x} \frac{t^2 y(x)-x^2 y(t)}{x-t}=3 $

Substituting

$ t \rightarrow x+h \Rightarrow h \rightarrow 0 $

$\lim\limits_{h \rightarrow 0} \frac{(x+h)^2 y(x)-x^2 y(x+h)}{-h}=3$

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

$\Rightarrow $ $\lim\limits_{h \rightarrow 0} \frac{x^2 y(x)+h^2 y(x)+2 x h y(x)-x^2 y(x+h)}{-h}=3$

$\Rightarrow $ $\lim\limits_{h \rightarrow 0} \frac{-x^2[y(x+h)-y(x)]+y(x)\left(h^2+2 x h\right)}{-h}=3$

$\Rightarrow $ $\lim\limits_{h \rightarrow 0} \frac{x^2[y(x+h)-y(x)]}{h}-\lim\limits_{h \rightarrow 0} \frac{y(x)(h+2 x) h}{h}=3$

Use

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

$ x^2 y^{\prime}(x)-2 x y(x)=3 $

Equation (1) is differential equation can be written by replacing

$ y^{\prime}(x)=\frac{d y}{d x} \quad \text { and } \quad y(x)=y $

$\Rightarrow $ $x^2 \frac{d y}{d x}-2 x y=3$

$\Rightarrow $ $\frac{d y}{d x}-\frac{2}{x} y=\frac{3}{x^2}$ (Linear differential equation)

Integrating factor

$ \text { I.F }=e^{\int-\frac{2}{x} d x}=e^{-2 \ln x}=e^{\ln \left(\frac{1}{x^2}\right)}=\frac{1}{x^2} $

$\therefore $ Solution of linear differential equation

$ y \cdot(\text { I.F })=\int(\text { I.F }) \cdot \frac{3}{x^2} d x $

$\Rightarrow $ $y \cdot \frac{1}{x^2}=\int \frac{1}{x^2} \cdot \frac{3}{x^2} d x=\int \frac{3}{x^4} d x$

$\Rightarrow $ $\frac{y}{x^2}=3 \times \frac{x^{-4+1}}{-4+1}=-\frac{1}{x^3}+C$

$\Rightarrow $ $ y=-\frac{1}{x}+C x^2 $ (General solution)

Apply initial condition

$ y(1)=2 $

$\Rightarrow $$ 2=-\frac{1}{1}+C(1) \Rightarrow C=3 $

So particular solution is : $ y(x)=-\frac{1}{x}+3 x^2$

$\therefore $ $y(2)=-\frac{1}{2}+3(2)^2=12-\frac{1}{2}=\frac{23}{2}$

$\Rightarrow $ $2 y(2)=23$

It is given that $f(x)$ is continuous at $x=0$

This mean at $x=0$

L.H.L = R.H.L = f(0) = a

Calculating L.H.L using the definition of $f(x)$ for $x<0$

$ \lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0} b^2 \sin \left(\frac{\pi}{2}\left[\frac{\pi}{2}(\cos x+\sin x) \cos x\right]\right)=a $

$\Rightarrow $ $\lim\limits_{x \rightarrow 0} b^2 \sin \left(\frac{\pi}{2}\left[\frac{\pi}{2}\left(\cos ^2 x+\frac{\sin 2 x}{2}\right)\right]\right)=a$

$\begin{gathered}\cos ^2\left(0^{-}\right)=1^{-} \text {Les than } 1 \\ \sin 2\left(0^{-}\right)=0^{-} \text {Less than } 0 \\ \Rightarrow a=b^2 \sin \left(\frac{\pi}{2}\left[\frac{3.14}{2}\left(1^{-}+0^{-}\right)\right]\right)\end{gathered}$

$\Rightarrow $ $a=b^2 \sin \left(\frac{\pi}{2}\left[1.57\left(1^{-}\right)\right]\right)$

$\Rightarrow $ $a=b^2 \sin \left(\frac{\pi}{2}\left[1.57^{-}\right]\right),\left[1.57^{-}\right]=1\{[t] \rightarrow$ G.I.F $\}$

$\Rightarrow $ $ a=b^2 \sin \left(\frac{\pi}{2} \times 1\right)=b^2 $

Calculating R.H.L using the function definition for $x>0$

$ \lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0^{+}} \frac{\sin x-\frac{1}{2} \sin 2 x}{x^3}=a $

$\Rightarrow $ $\lim\limits_{x \rightarrow 0^{+}} \frac{\sin x-\frac{1}{2} \times 2 \sin x \cos x}{x^3}=a$

$\Rightarrow $ $\lim\limits_{x \rightarrow 0^{+}} \frac{\sin x(1-\cos x)}{x \cdot x^2}=a$

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

Putting these result

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

Putting these result

$ a=1 \cdot \frac{1}{2}=\frac{1}{2} $

So $a^2+b^2=a^2+a\left\{b^2=a\right.$ from $\left.L . H . L\right\}$

$ =\frac{1}{4}+\frac{1}{2}=\frac{3}{4} $

$ f(x)= \begin{cases}2 \alpha\left(x^2-2\right)+2 \beta x, & x<1 \\ (\alpha+3) x+(\alpha-\beta), & x \geq 1\end{cases} $

Given $f(x)$ is differentiable at all $x \in \mathbb{R}$.

This means $f(x)$ is continuous and differentiable at $x=1$.

1. Continuity at $x=1$

$ \begin{aligned} & \text { L.H.L }=\text { R.H.L }=f(1) \\ & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ & \lim _{x \rightarrow 1^{-}}\left[2 \alpha\left(x^2-2\right)+2 \beta x\right]=\lim _{x \rightarrow 1^{+}}[(\alpha+3) x+(\alpha-\beta)] \\ & 2 \alpha(1-2)+2 \beta(1)=\alpha+3+\alpha-\beta \\ & -2 \alpha+2 \beta=2 \alpha-\beta+3 \\ & 3 \beta-4 \alpha=3-(1) \end{aligned} $

2. Differentiability at $x=1$

$ \begin{gathered} f^{\prime}(x)= \begin{cases}2 \alpha(2 x)+2 \beta, & x<1 \\ \alpha+3, & x>1\end{cases} \\ \text { L.H.D }=\text { R.H.D } \\ f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) \\ 2 \alpha(2 \cdot 1)+2 \beta=\alpha+3 \\ 3 \alpha+2 \beta=3-(2) \end{gathered} $

Solving equation (1) and (2)

$ \begin{gathered} \text { eqn }(1) \times 2-\text { eqn }(2) \times 3 \\ -8 \alpha-9 \alpha=-3 \Longrightarrow \alpha=\frac{3}{17} \end{gathered} $

Substitute $\alpha=\frac{3}{17}$ in equation (1):

$ \begin{gathered} 3 \beta-4 \times \frac{3}{17}=3 \\ 3 \beta=3+\frac{12}{17}=\frac{63}{17} \end{gathered} $

$ \beta=\frac{21}{17} $

Value of $34(\alpha+\beta)=34\left(\frac{3}{17}+\frac{21}{17}\right)=6+42=48$

Given that $f(x)$ is continuous at $x=0$ means $\lim\limits_{x \rightarrow 0} f(x)=f(0)$

$f(0)=\lim\limits_{x \rightarrow 0} \frac{e^x\left(e^{\tan x-x}-1\right)+\log _e(\sec x+\tan x)-x}{\tan x-x}$

$f(0)=\lim\limits_{x \rightarrow 0} \frac{e^x\left(e^{\tan x-x}-1\right)}{\tan x-x}+\lim\limits_{x \rightarrow 0} \frac{\log _e(\sec x+\tan x)-x}{\tan x-x}$

Using standard limit $\frac{e^{f(x)}-1}{f(x)}=1$ if $f(x) \rightarrow 0$

Let $f(x)=\tan x-x$, then $\lim\limits_{x \rightarrow 0} \frac{e^{\tan x-x}-1}{\tan x-x}=1$

Second limit is $\frac{0}{0}$ form use L'Hopital Rule

$ f(0)=1+\lim\limits_{x \rightarrow 0} \frac{\frac{d}{d x}(\log (\sec x+\tan x)-x)}{\frac{d}{d x}(\tan x-x)} $

$\begin{aligned} & =1+\lim _{x \rightarrow 0} \frac{\frac{1}{\sec x+\tan x}\left(\sec x \tan x+\sec ^2 x\right)-1}{\sec ^2 x-1} \\ & =1+\lim _{x \rightarrow 0} \frac{\sec x \tan x+\sec ^2 x-\sec x-\tan x}{(\sec x+\tan x)\left(\sec ^2 x-1\right)} \\ & =1+\lim _{x \rightarrow 0} \frac{\tan x(\sec x-1)+\sec x(\sec x-1)}{(\sec x+\tan x)\left(\sec ^2 x-1\right)} \\ & =1+\lim _{x \rightarrow 0} \frac{(\sec x-1)(\sec x+\tan x)}{(\sec x+\tan x)\left(\sec ^2 x-1\right)} \\ & =1+\lim _{x \rightarrow 0} \frac{1}{\sec x+1}\end{aligned}$

$\Rightarrow $ $f(0)=1+\frac{1}{2}=\frac{3}{2}$

Refunctioning the given function.

$ \begin{aligned} & f(x)= \begin{cases}\frac{a x+x^2-\sin 2 x}{x} & , x>0 \\ \frac{-a x+x^2+\sin 2 x}{x} & , x<0 \\ b & , x=0\end{cases} \\ & R H L=\lim _{x \rightarrow 0^{+}}\left(a+x-2 \frac{\sin 2 x}{2 x}\right)=a-2 \\ & L H L=\lim _{x \rightarrow 0^{-}}\left(-a+x+2 \frac{\sin 2 x}{2 x}\right)=-a+2 \end{aligned} $

For continuity:

$ \begin{aligned} & L H L=R H L \Rightarrow a-2=-a+2 \Rightarrow a=2 \\ & b=f(0)=R H L=2-2=0 \\ & a+b=2+0=2 \end{aligned} $

$ \begin{aligned} & \lim _{x \rightarrow \frac{-3}{2}} \frac{a x^2+2 a x+3}{4 x^2+4 x-3}=b \\ & a x^2+2 a x+3=0 \quad \text { at } n=\frac{-3}{2} \\ & \Rightarrow \frac{9 a}{4}-3 a+3=0 \\ & \Rightarrow \frac{3 a}{4}=3 \\ & \Rightarrow a=4 \\ & \lim _{x \rightarrow \frac{-3}{2}} \frac{4 x^2+8 a x+3}{4 x^2+4 x-3} \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \frac{-3}{2}} \frac{(2 x+1)(2 x+3)}{(2 x-1)(2 x+3)} \\ & =\frac{-2}{-4}=\frac{1}{2}=b \\ & \therefore f(x)=\frac{2 x+1}{2 x-1} \\ & f(f(x))=\frac{7}{5} \\ & \therefore \quad \frac{2 f(x)+1}{2 f(x)-1}=\frac{7}{5} \\ & \Rightarrow f(x)=3 \\ & \frac{2 x+1}{2 x-1}=3 \\ & \Rightarrow x=1 \end{aligned} $

$ \begin{aligned} & \lim _{x \rightarrow 0}\left[1+\frac{(a-1)}{1!} x+\frac{(a-1)^2}{2!} x^2+\ldots \ldots \ldots \infty\right] \\ & \frac{+2\left[1-\frac{(b x)^2}{2!}+\ldots \ldots \infty\right]+(c-2)\left[1-\frac{x^1}{1!}+\frac{x^2}{2!}-\ldots \ldots \infty\right]}{x\left(1-\frac{x^2}{2!}+\ldots \ldots \infty\right)-\left(x-\frac{x^2}{2}+\ldots \ldots \ldots \infty\right)} \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{(c+1)+(a-c+1) x+\left[\frac{(a-1)^2}{2}-b^2+\frac{(c-2)}{2}\right] x^2+\ldots \ldots \ldots \infty}{\left(\frac{x^2}{2}-\frac{x^3}{2}+\ldots \ldots \ldots \ldots \infty\right)}=2 \\ & \begin{array}{l} c+1=0 \Rightarrow c=-1 \\ a-c+1=0 \Rightarrow a+1+1=0 \Rightarrow a=-2 \\ \left.\frac{(a-1)^2}{2}-b^2+\frac{(c-2)}{2}\right) \\ \left(\frac{1}{2}\right) \\ \Rightarrow 9-2 b^2-3=2 \Rightarrow 2 b^2=4 \Rightarrow b^2=2 \\ \therefore a^2+b^2+c^2=4+2+1=7 \end{array} \end{aligned} $

$ \begin{aligned} &g(x)=|x|\left[x^2\right]\\ &\text { Points of discontinuity of } \mathrm{g}(\mathrm{x}) \text { in (-2,2) are }( \pm 1, \pm \sqrt{2}, \pm \sqrt{3})\\ &\begin{aligned} & \therefore S=\{-1,1,-\sqrt{2}, \sqrt{2} \cdot-\sqrt{3}, \sqrt{3}\} \\ & \because f(x)=\min \left\{\sqrt{2 x}, x^2\right\} \\ & \therefore \sum_{x \in S} f(x)=-\sqrt{2}+1-2+2-\sqrt{6}+\sqrt{6}=1-\sqrt{2} \end{aligned} \end{aligned} $

$ \ln \left(\lim\limits _{x \rightarrow 1}\left(\frac{f(x+2)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right)\left(1^{\infty}\right) $

$In\left( {{e^{\mathop {\lim }\limits_{x \to 1} }}{{18} \over {{{(x - 1)}^2}}}\left( {{{f(x + 2)} \over {f(3)}} - 1} \right)} \right)$

$In\left( {{e^{\mathop {\lim }\limits_{x \to 1} }}{{f'(x + 2)} \over {2(X - 1)}}} \right)$

$In\left( {{e^{\mathop {\lim }\limits_{x \to 1} }}{{f''(x + 2)} \over 2}} \right)$

$\mathop {\lim }\limits_{x \to 1} {{f''(3)} \over 2} = {4 \over 2} = 2$

(A)

Let

$f(x)=\begin{cases} \dfrac{1}{x}\sin\left(\dfrac{1}{x}\right), & x\neq 0\\\\ 0, & x=0 \end{cases}$

Now,

$g(x)=x f(x)=\begin{cases} \sin\left(\dfrac{1}{x}\right), & x\neq 0\\\\ 0, & x=0 \end{cases}$

When $x \neq 0$, the function $g(x)$ is $\sin\left(\dfrac{1}{x}\right)$, which keeps oscillating between $-1$ and $1$ as $x \to 0$. So, $\lim\limits_{x \to 0} g(x)$ does not exist. Therefore, $g(x)$ is not continuous at $x=0$.

(B)

Assume that $f(x)$ is continuous at $x=0$. This means $f(0^-) = f(0^+) = f(0)$.

Now consider the derivative of $g(x)$ at $x=0$:

$g'(0)=\lim\limits_{h\to 0}\dfrac{g(h)-g(0)}{h}$

Since $g(0)=0$, this becomes

$g'(0)=\lim\limits_{h\to 0}\dfrac{g(h)}{h}=\lim\limits_{h\to 0}\dfrac{h f(h)}{h}$

$g'(0)=\lim\limits_{h\to 0} f(h)=f(0)$

This limit exists and is finite because $f(x)$ is continuous at $x=0$. Hence, $g(x)$ is differentiable at $x=0$.

(C)

From the above, if $g(x)$ is differentiable at $x=0$, then

$g'(0)=\lim\limits_{h\to 0} f(h)$

This shows that the limit of $f(h)$ as $h \to 0$ exists. However, we cannot say anything about the value of $f(0)$. So, even if $g(x)$ is differentiable at $x=0$, we cannot conclude that $f(x)$ is continuous at $x=0$.

(D)

This statement is correct and follows from the reasoning in option (C).

Here $f(x)=\mathrm{e}^{\sin |x|}-|x|$. Since $|x|$ is not differentiable at $x=0$, and also $\sin|x|$ has a “kink” at $x=0$, the function $f$ is not differentiable at $x=0$. So, Statement I is false.

Now take $x \in\left(-\pi,-\frac{\pi}{2}\right)$. Here $x<0$, so $|x|=-x$ and hence $\sin|x|=\sin(-x)=-\sin x$. Therefore, on this interval we can write $f(x)=\mathrm{e}^{-\sin x}+x$ and differentiate to check whether $f$ is increasing.

$ \begin{aligned} & f(x)=e^{-\sin x}+x \\\\ & f^{\prime}(x)=-e^{-\sin x} \cos x+1 \\\\ & \text { for } x \in\left(-\pi, \frac{-\pi}{2}\right) \quad \cos x<0 \\\\ & \therefore f^{\prime}(x)>0 \\\\ & \therefore f(x) \text { is increasing } \end{aligned} $

Since $\cos x<0$ on $\left(-\pi,-\frac{\pi}{2}\right)$ and $e^{-\sin x}>0$, we get $-e^{-\sin x}\cos x>0$, so $f'(x)>0$ on this interval. Hence, Statement II is true.

$ \lim\limits_{x \rightarrow 0} \frac{x^2 \sin ^2 x}{x^2-\sin ^2 x} $

Applying expansion

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

$ \begin{aligned} & f(x)=\lim _{y \rightarrow 0} \frac{1-\cos (x y)}{(x y)^2} \cdot \frac{\tan x y}{x y} \frac{(x y)^3}{y^3} \\ & f(x)=\frac{x^3}{2} \\ & f(x)=\sin x \\ & \frac{x^3}{2}=\sin x \end{aligned} $

3 points of intersection

Let us define a new function $h(x)$ taking the difference between $f(x)$ and $g(x)$.

$h(x) = f(x) - g(x)$

Differentiating this function twice with respect to $x$, we get:

$h^{\prime \prime}(x) = f^{\prime \prime}(x) - g^{\prime \prime}(x)$

We are given that $f^{\prime \prime}(x) = g^{\prime \prime}(x)$ for all $x \in \mathbf{R}$. Therefore, their difference is zero:

$h^{\prime \prime}(x) = 0$

Now, let us integrate both sides with respect to $x$ to find $h^{\prime}(x)$:

$h^{\prime}(x) = \int 0 \,dx = c_1$

where $c_1$ is a constant of integration.

We are given the values of the first derivatives at $x = 1$. Let us simplify them:

$f^{\prime}(1) = 4$

$2g^{\prime}(1) = 4 \implies g^{\prime}(1) = 2$

Using these values, we can find the exact value of $h^{\prime}(1)$:

$h^{\prime}(1) = f^{\prime}(1) - g^{\prime}(1) = 4 - 2 = 2$

Since $h^{\prime}(x) = c_1$ for all values of $x$, it follows that:

$c_1 = 2$

Substituting this back, we get our first derivative equation:

$h^{\prime}(x) = 2$

Next, we need the original function $h(x)$. We find it by integrating $h^{\prime}(x)$ with respect to $x$:

$h(x) = \int 2 \,dx = 2x + c_2$

where $c_2$ is another constant of integration.

We are given the values of the functions at $x = 2$. Let us simplify these values as well:

$g(2) = 9$

$3f(2) = 9 \implies f(2) = 3$

Using these values, we can find the value of $h(2)$:

$h(2) = f(2) - g(2) = 3 - 9 = -6$

Now, substitute $x = 2$ and $h(2) = -6$ into our integrated equation for $h(x)$:

$h(2) = 2(2) + c_2$

$-6 = 4 + c_2$

$c_2 = -6 - 4$

$c_2 = -10$

Substituting the value of $c_2$ back into the equation for $h(x)$, we have the completed function rule:

$h(x) = 2x - 10$

The question asks for the value of $f(25) - g(25)$, which is represented directly by $h(25)$. Let us substitute $x = 25$ into our function $h(x)$:

$h(25) = 2(25) - 10$

$h(25) = 50 - 10$

$h(25) = 40$

Therefore, $f(25) - g(25) = 40$.

The correct answer is Option B.

$ \lim\limits_{x \rightarrow 0} \frac{1-\cos \alpha x \cdot \cos ((\alpha+1) x) \cos ((\alpha+2) x)}{(\alpha+1)^2 x^2} $

By using L.H. Rule we get

$ \begin{aligned} & \Rightarrow \frac{1}{2}(\alpha+1)^2\left[\alpha^2+(\alpha+1)^2+(\alpha+2)^2\right]=2 \\ & \alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2 \\ & \alpha^2+(\alpha+2)^2=3(\alpha+1)^2 \\ & \Rightarrow \alpha^2+2 \alpha-1=0 \\ & \text { Product }=-1 \end{aligned} $

$\therefore$ Denominator $=0$ at $x=2$

$\Rightarrow$ Numerator $=0$ at $x=2$

$2^3-5(2)^2+a(2)+b=0$

$\Rightarrow 2a+b=12 \quad \ldots (1)$

$\therefore \; m=\lim\limits_{x\rightarrow 2}\frac{\dfrac{\sin\left(x^3-5x^2+ax+b\right)}{x^3-5x^2+ax+b}\cdot \left(x^3-5x^2+ax+b\right)}{\left(\dfrac{x-1-1}{\sqrt{x-1}+1}\right)\cdot \dfrac{\log_e\left(1+(x-2)\right)}{x-2}\cdot (x-2)}$

$\Rightarrow m=\lim\limits_{x\rightarrow 2} 2\left(\frac{x^3-5x^2+ax+b}{(x-2)^2}\right)$

$\Rightarrow m=\lim\limits_{x\rightarrow 2} 2\left(\frac{3x^2-10x+a}{2(x-2)}\right)$

$\therefore$ Denominator $=0$ at $x=2$

$\Rightarrow$ Numerator $=0$ at $x=2$

$\Rightarrow 3(2)^2-10(2)+a=0 \rightarrow a=8$

put $a=8$ in (1), we get $b=-4$

$\therefore \; m=\lim\limits_{x\rightarrow 2} 2\left(\frac{6x-10}{2}\right) \rightarrow m=2$

$\therefore \; a+b+m=6$

$\begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots .-2 x}{x^5}=\frac{2}{5} \\ & \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow 1}\left(\frac{2}{1-x}\right)(x-1)}=e^{-2} \end{aligned}\\ &\Rightarrow \text { Both statements correct } \end{aligned}$

$\lim _{x \rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \ln \left(1+3 x^3\right)}{\left(\tan ^{-1}(3 \sqrt{x})\right)^2\left(e^{5 x^{\frac{4}{3}}}-1\right)}$

$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^2})} \over {3{x^2}}} \times 5{{(x)}^{{1 \over 3}}}(3{x^2})} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5x{4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 9x \times 5{x^{{4 \over 3}}}}}$

$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^3})} \over {3{x^2}}} \times 15{x^{{7 \over 3}}}} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5(x){4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 45{x^{{7 \over 3}}}}}$

$ = {1 \over 3}$

$\begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0}(f(2+x))^{3 / x}=\left(1^{\infty} \text { form }\right) \\ & e^{\lim _{x \rightarrow 0 x}^3(f(2+x)-1)}=e^{\lim _{x \rightarrow} 3 f^{\prime}(2+x)} \\ & =e^{3 f^{\prime}(2)} \\ & =e^{12} \\ & \Rightarrow \alpha=12 \\ & y=4 x^3-4 x^2-4(12-7) x-12 \\ & y=4 x^3-4 x^2-20 x-12 \\ & y=4\left(x^3-x^2-5 x-3\right) \\ & =4(x+1)^2(x-3) \end{aligned}\\ &\text { It meets the } x \text {-axis at two points } \end{aligned}$

$\begin{aligned} & f(0)=1, f(2 x)-f(x)=x \\ & \text { Replace } x \rightarrow \frac{x}{2} \\ & f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}\quad\text{..... (1)} \end{aligned}$

Again, Replace $x \rightarrow \frac{x}{2}$

$f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^2}\right)=\frac{x}{2^2}\quad\text{...... (2)}$

: $\qquad$ : $\qquad$ :

: $\qquad$ : $\qquad$ :

: $\qquad$ : $\qquad$ :

$\begin{aligned} & f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \\ & \text { Adding }(1)+(2)+(3)+\ldots+(\mathrm{n}) \\ & \text { We get } f(x)-f\left(\frac{x}{2^n}\right)=\frac{x}{2}+\frac{x}{2^2}+\ldots .+\frac{x}{2^n} \\ & \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{x}{2}+\frac{x}{2^n}+\ldots+\frac{x}{2^n}\right) \end{aligned}$

$\begin{aligned} & f(x)-f(0)=\frac{\frac{x}{2}}{\frac{1}{2}} \\ & \Rightarrow G(x)=x \\ & \Rightarrow G\left(r^2\right)=r^2 \\ & \Rightarrow \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} G\left(r^2\right) \\ & =\frac{(10)(11)(21)}{6}=(55) 7 \\ & \Rightarrow 385 \end{aligned}$

$\begin{aligned} &\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1\\ &\text { Let } x-1=t\\ &\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t \cos t-\mu \sin t}{t^3}=-1 \end{aligned}$

$\begin{aligned} & =\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t\left(1-\frac{t^2}{2!}+\frac{t^4}{4!}+\cdots\right)-\mu\left(t-\frac{t^3}{3!}+\cdots\right)}{t^3}=-1 \\ & =\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda-\mu)+t^3\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)+\cdots}{t^3}=-1 \end{aligned}$

$\begin{aligned} &\begin{aligned} \therefore\quad & \lambda-\mu+6=0 \quad\text{..... (i)}\\ & \frac{\mu}{6}-\frac{\lambda}{2}=-1 \quad\text{..... (ii)} \end{aligned}\\ &\text { Solving (i) and (ii) }\\ &\begin{aligned} & \lambda=6, \mu=12 \\ & \lambda+\mu=18 \end{aligned} \end{aligned}$

$f(x)= \begin{cases}(1+a x)^{1 / x} & , x<0 \\ 1+b & , x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} & , x>0\end{cases}$

$\begin{aligned} &\text { Hence, } f(0)=1+6\\ &\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2} \quad\left[\text { For } \frac{0}{0} \text { form, } c=8\right] \end{aligned}$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{2\left(1+\frac{x}{4}\right)^{1 / 2}-2}{\left(1+\frac{x}{8}\right)^{1 / 3}-2} \\ & =\lim _{x \rightarrow 0} \frac{1+\frac{x}{8}-1}{1+\frac{x}{8} \cdot \frac{1}{3}-1}=\frac{\frac{1}{8}}{\frac{1}{8} \cdot \frac{1}{3}}=3 \\ & \therefore 1+b=3 \\ & \Rightarrow b=2 \end{aligned}$

$\begin{aligned} & \mathrm{LHL}=\lim _{x \rightarrow 0}(1+a x)^{1 / x}=\lim _{x \rightarrow 0} e^{\frac{1+a x-1}{x}}=e^a=3 \\ & \therefore \quad e^a \cdot b c=3 \cdot 2.8=48 \end{aligned}$

To find the value of $a + b$ for which the following limit is finite:

$ \lim\limits_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} $

we start by expanding the cosine functions using their Taylor series:

$ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \ldots $

$ \cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldots $

Substitute these expansions into the limit expression:

$ \lim\limits_{x \to 0} \frac{\left(1 - \frac{4x^2}{2!} + \frac{(2x)^4}{4!} - \ldots \right) + a\left(1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldots\right) - b}{x^4} $

To ensure that the limit is finite, the linear and quadratic terms must sum to zero. Let's equate the constant terms first:

For the limit to be finite:

$ 1 + a - b = 0 $

Equating the coefficients of $x^2$:

$ -2 - 8a = 0 $

Solving the second equation, we have:

$ 8a = -2 \\ a = \frac{-1}{4} $

Substitute $a$ back into the first equation to find $b$:

$ 1 + \frac{-1}{4} - b = 0 \\ b = 1 - \frac{1}{4} \\ b = \frac{3}{4} $

Thus, the sum of $a$ and $b$ is:

$ a + b = \frac{-1}{4} + \frac{3}{4} = \frac{1}{2} $

Therefore, the correct value of $a + b$ is $\frac{1}{2}$.

$\begin{aligned} & \text { At } x \rightarrow 0 \\ & \sin 2 x-\beta x \rightarrow 0 \\ & \Rightarrow \quad \frac{0}{0} \text { form } \\ & \Rightarrow \quad(\gamma-1) e^0+0 \sin (\alpha x) \rightarrow 0 \\ & \Rightarrow \quad(\gamma-1)=0 \\ & \Rightarrow \quad \gamma=1 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{x^2 \sin (\alpha x)}{(\sin 2 x-\beta x)}=3 \end{aligned}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{x^2\left[\alpha x-\frac{(\alpha x)^3}{3!}+\frac{(\alpha x)^5}{5!}-\cdots\right]}{\left[(2 x)-\frac{(2 x)^3}{3!}+\frac{(2 x)^5}{5!}\right]-\beta x}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{\alpha x^3-\frac{\alpha^3 x^5}{3!}+\frac{\alpha^5 x^7}{5!}-\cdots}{x(2-\beta)-\frac{8 x^3}{6}+\frac{2^5 \cdot x^5}{5!}-\cdots}=3$

$\begin{aligned} \Rightarrow & 2-\beta=0 \text { and } \frac{\alpha}{\frac{-8}{6}}=3 \\ \Rightarrow & \beta=2 \\ & \alpha=3\left(-\frac{8}{6}\right)=-4 \\ \Rightarrow & \gamma=1, \beta=2, \alpha=-4 \\ \Rightarrow & \beta+\gamma-\alpha=7 \end{aligned}$

$\cos |\mathrm{x}|$ is always differentiable

$\therefore$ we have to check only for $\left|\mathrm{x}^2-\mathrm{ax}+2\right|$

$\therefore$ Not differentiable at

$x^2-a x+2=0$

One root is given, $\alpha=2$

$\begin{aligned} \therefore \quad 4 & -2 a+2=0 \\ & a=3 \end{aligned}$

$\therefore$ other root $\beta=1$

but for $x=1 f(x)$ is differentiable

(Drop)

$\begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+6-1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right) \end{aligned}$

$\begin{aligned} & =\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} \ldots+\frac{1}{\mathrm{n!}}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!} \ldots-\frac{1}{(\mathrm{n}+3)!}\right) \\ & \quad=\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3} \end{aligned}$

$f(x)=[x]+|x-2| \quad-2< x<3$

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

So $f(x)$ is not continuous at 4 points and not differentiable at 4 point

So $\mathrm{m}+\mathrm{n}=4+4=8$

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

$F(x)-6 f(1 / x)=\frac{35}{3 x}-\frac{5}{2}\quad\text{..... (1)}$

$\begin{aligned} &\begin{aligned} & \text { Replace } x \rightarrow \frac{1}{x} \\ & F(1 / x)-6(x)=\frac{35 x}{3}-\frac{5}{2}\quad\text{..... (2)} \end{aligned}\\ &\text { Using (1) & (2) }\\ &\begin{aligned} & f(x)=-2 x-\frac{1}{3 x}+\frac{1}{2} \\ & B=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right) \end{aligned} \end{aligned}$

$\begin{aligned} &= \lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}-2 x-\frac{1}{3 x}+\frac{1}{2}\right) \\ & \alpha=3, \quad B=1 / 2 \\ & \text { So, } \alpha+2 B=3+1=4 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right)\left(1-\frac{1}{3 x}\right)^{x / 2}}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right)\left(1+\frac{2}{3 x}\right)^{x / 2}} \\ & =\lim _{x \rightarrow \infty} \frac{2}{3} \cdot \frac{e^{\frac{x}{2}\left(1-\frac{1}{3 x}-1\right)}}{e^{\frac{x}{2}\left(1+\frac{2}{3 x}-1\right)}} \\ & =\frac{2}{3} \cdot \frac{e^{-\frac{1}{6}}}{e^{1 / 3}}=\frac{2}{3} e^{-\frac{1}{2}} \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\ & \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\ & \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=4 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2\right) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=2 \\ & \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \\ & \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \\ & \therefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \\ & \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10 \end{aligned}$

$ \begin{aligned} \alpha &= \lim_{x \to \infty} \left[\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right]^x. \end{aligned} $

Begin by rewriting the expression inside the limit. Notice that

$ \frac{x}{1+x} = 1 - \frac{1}{1+x}, $

so

$ \frac{1}{e} - \frac{x}{1+x} = \frac{1}{e} - \left(1 - \frac{1}{1+x}\right) = \frac{1}{e} - 1 + \frac{1}{1+x}. $

Express the constant part as

$ 1 - \frac{1}{e} = \frac{e-1}{e}, $

to obtain

$ \frac{1}{e} - \frac{x}{1+x} = -\frac{e-1}{e} + \frac{1}{1+x}. $

Multiplying by the prefactor, we have

$ \left(\frac{e}{1-e}\right)\left(\frac{1}{e} - \frac{x}{1+x}\right) = \frac{e}{1-e}\left[-\frac{e-1}{e} + \frac{1}{1+x}\right]. $

Split the expression:

$ \frac{e}{1-e}\cdot\left(-\frac{e-1}{e}\right) + \frac{e}{1-e}\cdot\frac{1}{1+x} = -\frac{e-1}{1-e} + \frac{e}{(1-e)(1+x)}. $

Since

$ 1-e = -(e-1), $

it follows that

$ -\frac{e-1}{1-e} = -\frac{e-1}{-(e-1)} = 1. $

Hence, the expression simplifies to

$ 1 + \frac{e}{(1-e)(1+x)}. $

Again, replacing $1-e$ by $-(e-1)$,

$ 1 + \frac{e}{-(e-1)(1+x)} = 1 - \frac{e}{(e-1)(1+x)}. $

Thus, the limit becomes

$ \alpha = \lim_{x \to \infty} \left(1 - \frac{e}{(e-1)(1+x)}\right)^x. $

For large $x$, note that

$ 1+x \sim x, $

so we approximate

$ \alpha = \lim_{x \to \infty} \left(1 - \frac{e}{(e-1)x}\right)^x. $

Recall the limit

$ \lim_{x \to \infty} \left(1 - \frac{a}{x}\right)^x = e^{-a}, $

with

$ a = \frac{e}{e-1}. $

Therefore,

$ \alpha = e^{-\frac{e}{e-1}}. $

Taking the natural logarithm,

$ \ln \alpha = -\frac{e}{e-1}. $

Now, compute

$ \frac{\ln \alpha}{1 + \ln \alpha} = \frac{-\frac{e}{e-1}}{1 - \frac{e}{e-1}}. $

Express the denominator with a common denominator:

$ 1 - \frac{e}{e-1} = \frac{e-1}{e-1} - \frac{e}{e-1} = -\frac{1}{e-1}. $

Thus,

$ \frac{-\frac{e}{e-1}}{-\frac{1}{e-1}} = \frac{e}{1} = e. $

The final result is

$ \frac{\ln\alpha}{1+\ln\alpha} = e. $

$\begin{aligned} & T_n=S_n-S_{n-1} \\\\ & \Rightarrow T_n=\frac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3) \\\\ & \Rightarrow \frac{1}{T_{\mathrm{n}}}=\frac{8}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\\\ & \lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{T_{\mathrm{r}}}=\lim _{\mathrm{n} \rightarrow \infty} 8 \sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\\\ & =\lim _{\mathrm{n} \rightarrow \infty} \frac{8}{4} \sum\left(\frac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)}-\frac{1}{(2 \mathrm{n}+1)(2 \mathrm{n}+3)}\right) \\\\ & =\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\ldots\right] \\\\ & =\frac{2}{3}\end{aligned}$