Question

$\underset{x\to 0}{lim}\frac{x}{{\tan }^{-1}2x}$ is equal to

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $\infty$

Answer 1

The correct option is B $\frac{1}{2}$

Substituting we get indeterminate form,
So, applying L'Hospital's Rule,
$={lim}_{x\to 0}\frac{1}{\left(\frac{2}{(1+4x^2)}\right)}$
$=\frac{1}{2}$
Hence, option 'B' is correct.