Question

If $f(x)=\frac{{\log }_e(1+x^2\tan x)}{\sin x^3},x\ne 0$ is to be continuous at $x=0,$ then $f(0)$ must be defined as

• A. $1$
• B. $0$
• C. $1/2$
• D. $-1$
• E. $2$

Answer 1

Answer: (A)

$f(x)=\frac{{\log }_e(1+x^2\tan x)}{\sin x^3}$ This function is continuous at $x=0,$ then $\underset{x\to 0}{\lim }\frac{{\log }_e(1+x^2\tan x)}{\sin x^3}=f(0)$
$\Rightarrow$ $\underset{x\to 0}{\lim }\frac{{\log }_e\left\{1+x^2\left(x+\frac{x^3}{3}+\frac{2x^5}{15}+.....\right)\right\}}{x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-....}$
$\Rightarrow$ $\underset{x\to 0}{\lim }\frac{{\log }_e(1+x^3)}{x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-....}=f(0)$ [on neglecting higher power of $x$ in $x^{2}tanx$ ]
$\Rightarrow$ $\underset{x\to 0}{\lim }\frac{x^3-\frac{x^6}{2}+\frac{x^9}{3}-.....}{x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-....}=f(0)$
$\Rightarrow$ $1=f(0)$