Question

If $L=\underset{x\to \frac{\pi }{4}}{lim}(2-\tan x{)}^{\frac{1}{\ln \left(\tan x\right)}}$, then find $L$.

• A. $e^{-1}$
• B. $e$
• C. $e^2$
• D. $e^{-2}$

Answer 1

Answer: (A)

$e^{-1}$

$\underset{x\to \frac{\pi }{4}}{lim}{(2-\tan x)}^{\frac{1}{\ln }\left(\tan x\right)}=e^{\underset{x\to \frac{\pi }{4}}{lim}\frac{\ln (2-\tan x)}{\ln \left(\tan x\right)}}$
By applying L'Hospital rule
$=e^{\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{-{\sec }^{2}x}{2-\tan x}}{\frac{{\sec }^{2}x}{\tan x}}}=e^{\underset{x\to \frac{\pi }{4}}{lim}\frac{\tan x}{\tan x-2}}=e^{-1}$