#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 34 Maths Textbook Solution.

$f\left ( 0 \right )=1$

Hint:

$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.

Given:

$f(x)=\frac{2 x+3 \sin x}{3 x+2 \sin x}, x \neq 0$

Solution:

\begin{aligned} &f(x)=\frac{2 x+3 \sin x}{3 x+2 \sin x}, x \neq 0 \\\\ &\lim _{x \rightarrow 0} f(x)=f(0)\\ \\ &\lim _{x \rightarrow 0} \frac{2 x+3 \sin x}{3 x+2 \sin x}=f(0) \\\\ &\lim _{x \rightarrow 0} \frac{x\left(2+3 \frac{\sin x}{x}\right)}{x\left(3+2 \frac{\sin x}{x}\right)}=f(0) \end{aligned}

\begin{aligned} &\lim _{x \rightarrow 0} \frac{\left(2+3 \frac{\sin x}{x}\right)}{\left(3+2 \frac{\sin x}{x}\right)}=f(0) \\\\ &\frac{\lim _{x \rightarrow 0}\left(2+3 \frac{\sin x}{x}\right)}{\lim _{x \rightarrow 0}\left(3+2 \frac{\sin x}{x}\right)}=f(0) \end{aligned}

$\frac{(2+3) \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)}{(3+2) \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)}=f(0)$                                                                                       $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

\begin{aligned} &\frac{(2+3) 1}{(3+2) 1}=f(0) \\ &\frac{5}{5}=f(0) \\ &f(0)=1 \end{aligned}