#### Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 7

The correct option is (C)

Hint:

Use the given formula:

$\text { (i) } \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\tan x}{x}=1$

(ii)  A function f(x) is said to be continuous at a point x = a of its domain

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text {iii}) \lim _{x \rightarrow a}\{f(x) g(x)\}=1 m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}

Given:

$f(x)=(x+1)^{\cot x}$

Solution:

$f(x)=(x+1)^{\cot x}$

taking log on both sides

$\log f(x)=(\cot x)(\log (x+1))$

$\lim _{x \rightarrow 0} \log f(x)=\lim _{x \rightarrow 0}\left(\frac{\frac{\log (x+1)}{x}}{\frac{\tan x}{x}}\right)$

Using formula (iii)

$\lim _{x \rightarrow 0} \log f(x)=\frac{\lim _{x \rightarrow 0} \frac{\log (x+1)}{x}}{\lim _{x \rightarrow 0} \frac{\tan x}{x}}$

Using standard limit formula (i)

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=e \\ &f(0)=e \end{aligned}

So, the correct option is (C)