Question

Evaluate the limit .$\lim_{x \to \pi/2 }$$(tanx)^{(\pi/2)-x}$using L’Hôpital’s rule

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \lim_{x \to \frac{\pi}{2} } \: \big( \tan(x)\big)^{ \frac{\pi}{2} - x }$

$\tt{Put \: \: \: \dfrac{\pi}{2} - x= t}$

$\displaystyle \tt{ = \lim_{t \to0} \: \left \{ tan \left( \dfrac{\pi}{2} - t\right)\right \}^{ t} }$

$\displaystyle \tt{ = \lim_{t \to0} \: \left \{ cot \left( t\right)\right \}^{ t} }$

$\displaystyle \tt{ = \lim_{t \to0} \: {e}^{ \displaystyle \: \tt{ t\ln(cot ( t))}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ \lim_{t \to0}t\ln(cot ( t))}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ \lim_{t \to0} \dfrac{\ln(cot ( t))}{ \frac{1}{t} }}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ \lim_{t \to0} \dfrac{ - cosec^{2} ( t)}{ - cot(t) \cdot\frac{1}{t ^{2} } }}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ \lim_{t \to0} \dfrac{ t^{2} }{ sin(t) cos(t) }}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ \lim_{t \to0} \dfrac{ t }{ sin(t) } \cdot \lim_{t \to0} \dfrac{ t }{ cos(t) }}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ 1 \cdot \lim_{t \to0} \dfrac{ 1 }{ - sin(t) }}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ - \lim_{t \to0} \dfrac{ 1 }{ sin(t) }}} }$

$\displaystyle \tt{ = {e}^{ \displaystyle \: \tt{ - \infty }} }$

$\displaystyle \tt{ = 0 }$