Question

The value of f(0) so that the function f(x) = $\frac{log(1+x^{2}tanx)}{sin{x}^3}$, x = 0 continuous at x = 0, is

• A. 1
• B. 0
• C. 2
• D. -1

Answer 1

The correct option is A 1

At $x=0f\left(0\right)=\frac{0}{0}$ [Indeterminate form]

So, applying L Hospital rule we get

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

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

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

[Since $\underset{x\to 0}{lim}\frac{\tan x}{x}=1$ & $\underset{x\to 0}{lim}{\sec }^{2}x=1$]

Therefore substituting $x=0$, we get $\Rightarrow$

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

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