Solve this

Given:

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

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 x+3 \sin x}{3 x+2 \sin x}=f(0)$

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

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

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

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

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

$\Rightarrow \frac{5}{5}=f(0)$

$\Rightarrow f(0)=1$