If f x = log e 1 + x 2 ( tan x ) sin x 3 , x ≠ 0 is to be continuous at x = 0 , then f ( 0 ) must be…

$f\left(x\right)=\frac{{\log }_e\left(1+x^2\left(\tan x\right)\right)}{\sin x^3},x\ne 0$

$f\left(0\right)=\frac{{\log }_e\left(1+0\left(\tan 0\right)\right)}{\sin 0}=\frac{0}{0}$

Apply $\text{L'Hospital}$ rule

$\frac{d\left(\mathrm{uv}\right)}{d\mathrm{x}}={\mathrm{uv}}^{\text{'}}+{\mathrm{vu}}^{\text{'}}$

$\Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }\frac{1}{1+x^2\tan x}\left(\frac{x^2{\sec }^{2}x+2x\tan x}{\cos x^3\times 3x^2}\right)$

$\Rightarrow f\left(0\right)=\underset{x\to 0(}{\lim }\frac{x}{1+x^2\tan x)}\frac{(x{\sec }^{2}x+2\tan x)}{\cos x^3\times 3x^2}$

$=\underset{x\to 0}{\lim }\frac{xse{c}^{2}x+2\tan x}{3x\cos x^3\left(1+x^2\tan x\right)}$

$=\underset{x\to 0}{\lim }\frac{2{\displaystyle \frac{\tan x}{x}}+se{c}^{2}x}{3\cos x^3\left(1+x^2\tan x\right)}$

$\because \underset{x\to 0}{\lim }\frac{\tan x}{x}=1$

$=\underset{x\to 0}{\lim }\frac{2\times 1+1}{3\times 1\left(1+0\right)}=\frac{3}{3}=1$.