Question

Find the value of $ {\lim }_{x \to \frac{\pi }{2}} {\left( {sinx} \right)^{\tan x}}$ .
A. 1
B. 0
C. e
D. None of these

Answer 1

Hint: First suppose that the given expression is equal to y without the limit part, then take logarithm to both sides of the obtained equation. After that, express $\tan x$ as $\dfrac{{\sin x}}{{\cos x}}$ . Then apply L’Hopital’s rule to obtain the required result.

Formula Used:
${\log _a}b = x \Rightarrow b = {a^x}$
The L’Hopital’s rule states that for $ {\lim }_{x \to a} \dfrac{{f(x)}}{{g(x)}}$ takes the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ then we can write the given expression as $ {\lim }_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$.

Complete step by step solution:
Suppose that,
$y = {\left( {sinx} \right)^{\tan x}}$
Now, taking logarithm to both sides we have,
$\log y = \log {\left( {sinx} \right)^{\tan x}}$
$ = \tan x\log \left( {sinx} \right)$
$ = \dfrac{{sinx.\log (\sin x)}}{{\cos x}}$
Taking the limit of x as x tending to $\dfrac{\pi }{2}$ .
$ {\lim }_{x \to \frac{\pi }{2}} \log y = {\lim }_{x \to \frac{\pi }{2}} \left( {\dfrac{{sinx.\log (\sin x)}}{{\cos x}}} \right)$
$ = \dfrac{{ {\lim }_{x \to \frac{\pi }{2}} \left( {sinx.\log (\sin x)} \right)}}{{ {\lim }_{x \to \frac{\pi }{2}} \cos x}}\left( {\dfrac{0}{0}} \right)$
$ = \dfrac{{ {\lim }_{x \to \frac{\pi }{2}} \left( {\cos x.\log (\sin x) + \sin x.\dfrac{1}{{\sin x}}.\cos x} \right)}}{{ {\lim }_{x \to \frac{\pi }{2}} ( - \sin x)}}$
=0
Therefore, $\log \left( { {\lim }_{x \to \frac{\pi }{2}} y} \right) = 0$
$ \Rightarrow {\lim }_{x \to \frac{\pi }{2}} y = {e^0} = 1$

Option ‘A’ is correct

Note:We often tend to substitute $\dfrac{\pi }{2}$ for x in the given expression to obtain the answer but that is not the correct way to answer this question. First suppose the expression as y then take log and calculate to obtain the correct answer.