Question

${lim}_{x\to \frac{\pi }{2}}(1-sinx)tanx$ will be equal to _____

• A. 1
• B. -1
• C. 0
• D. None of the above

Answer 1

The correct option is C

0

$Asx\to \frac{\pi }{2},sinx\to 1$

$i.e.(1-sinx)\to 0and$

$asx\to \frac{\pi }{2},tanx\to \infty$

$or\frac{1}{tanx}=cotx\to 0$

So using this information we can say that $x\to \frac{\pi }{2},(1-sinx)tanxor\frac{(1-sinx)}{cotx}comesto\frac{0}{0}form$

So we can apply L-Hospitals rule.

${lim}_{x\to \frac{\pi }{2}}\frac{(1-sinx)}{cotx}={lim}_{x\to \frac{\pi }{2}}\frac{\frac{d}{dx}(1-sinx)}{\frac{d}{dx}\left(cotx\right)}={lim}_{x\to \frac{\pi }{2}}\frac{(-cosx)}{-cose{c}^{2}x}$

$={lim}_{x\to \frac{\pi }{2}}si{n}^{2}xcosx=0$

Which is the correct answer.