Question

The value of the limit $\underset{x\to 0}{lim}(\frac{x}{\sin x}{)}^{\frac{6}{x^2}}$ is

• A. $e$
• B. $e^{-1}$
• C. $e^{-\frac{1}{6}}$
• D. $e^6$

Answer 1

The correct option is A $e$

$\underset{x\to 0}{lim}(\frac{x}{\sin x}{)}^{\frac{6}{x^2}}\Rightarrow (1{)}^{\infty }\text{form}$
$\therefore e^{\underset{x\to 0}{lim}\frac{6}{x^2}(\frac{x}{\sin x}-1)}$
$e^{\underset{x\to 0}{lim}\frac{6}{x^2}(\frac{x-\sin x}{\sin x})}$
$=e^{\underset{x\to 0}{lim}\frac{6}{x^2}\left(\frac{x-(x-\frac{x^3}{3!}+\frac{x^5}{5!}.....)}{(x-\frac{x^3}{3!}+\frac{x^5}{5!}.....)}\right)}$
$=e^{\underset{x\to 0}{lim}\frac{6}{x^2}\left(\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}.....)}{(x-\frac{x^3}{3!}+\frac{x^5}{5!}.....)}\right)}$
$=e^{\underset{x\to 0}{lim}\frac{6x^3}{x^3}\left(\frac{(\frac{1}{3!}-\frac{x^2}{5!}.....)}{(1-\frac{x^2}{3!}+\frac{x^4}{5!}.....)}\right)}=e^{\frac{6}{3!}}=e^1=e$