Question

$\underset{x\to \infty }{lim}\frac{e^{\frac{1}{x^2}}-1}{2{\tan }^{-1}\left(x^2\right)-\pi }$ is equal to

• A. $1$
• B. $-1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D $-\frac{1}{2}$

$\underset{x\to \infty }{lim}\frac{e^{\frac{1}{x^2}}-1}{2{\tan }^{-1}(x{)}^2-\pi }$
Put $t=\frac{1}{x}$
As $x\to \infty \Rightarrow t\to 0$
$RHL=\underset{t\to 0^{+}}{lim}\frac{e^{t^2}-1}{2{\cot }^{-1}{t}^2-\pi }$
$=\underset{t\to 0^{+}}{lim}\frac{e^{t^2}-1}{2(\frac{\pi }{2}-{\tan }^{-1}{t}^2)-\pi }$
$=\underset{t\to 0^{+}}{lim}\frac{e^{t^2}-1}{-2{\tan }^{-1}{t}^2}$
$=\underset{t\to 0^{+}}{lim}-\frac{1}{2}\frac{e^{t^2}-1}{t^2\frac{\tan t^2}{t^2}}=-\frac{1}{2}$