Question

$\underset{x\to 0}{lim}{\left(\begin{array}{c}\frac{3}{1+\sqrt{4+x}}\end{array}\right)}^{cosecx}$ has the value equal to -

• A. $e^{-1/12}$
• B. $e^{-1/6}$
• C. $e^{-1/4}$
• D. $e^{-1/3}$

Answer 1

Answer: (A)

$e^{-1/12}$

$\underset{x\to 0}{lim}{\left(\frac{3}{1+\sqrt{x+4}}\right)}^{\frac{1}{\sin x}}=\underset{x\to 0}{lim}(3^{\frac{1}{\sin x}}\times {\left(\frac{1}{1+\sqrt{x+4}}\right)}^{\frac{1}{\sin x}})$
$=exp\left(\underset{x\to 0}{lim}(\frac{\log 3}{\sin x}+\frac{\log \left(\frac{1}{1+\sqrt{x+4}}\right)}{\sin x})\right)$
$=exp\left(\underset{x\to 0}{lim}\left(\csc (\log 3+\log \left(\frac{1}{1+\sqrt{x+4}}\right))\right)\right)=e^{-\frac{1}{12}}$