Question

If $f\left(x\right)=\frac{2x+3\sin x}{3x+2\sin x},$ x ≠ 0 is continuous at x = 0, then find f (0).

Answer 1

Given:
$f\left(x\right)=\frac{2x+3\sin x}{3x+2\sin x},x\ne 0$

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

​$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{2x+3\sin x}{3x+2\sin x}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{x\left(2+3{\displaystyle \frac{\sin x}{x}}\right)}{x\left(3+2{\displaystyle \frac{\sin x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{\left(2+3{\displaystyle \frac{\sin x}{x}}\right)}{\left(3+2{\displaystyle \frac{\sin x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \frac{\underset{x\to 0}{\lim }\left(2+3{\displaystyle \frac{\sin x}{x}}\right)}{\underset{x\to 0}{\lim }\left(3+2{\displaystyle \frac{\sin x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \frac{2+3\underset{x\to 0}{\lim }\left({\displaystyle \frac{\sin x}{x}}\right)}{3+2\underset{x\to 0}{\lim }\left({\displaystyle \frac{\sin x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \frac{2+3\times 1}{3+2\times 1}=f\left(0\right) \\ \Rightarrow \frac{5}{5}=f\left(0\right) \\ \Rightarrow f\left(0\right)=1 \end{gathered}$$