Question

$\displaystyle \lim_{x\to\pi/ 4} \frac{\tan\,x-1}{\cos\,2x}$ is equal to

• A. $1$
• B. $0$
• C. $-2$
• D. $-1$

Answer 1

Answer: (D)

$\displaystyle\lim_{x\to\pi/ 4} \frac{\tan\,x-1}{\cos\,2x}$
$ = \displaystyle\lim_{h\to0} \frac{\tan\left(\frac{\pi}{4}+h\right)-1}{\cos\,2\left(\frac{\pi }{4}+h\right)}$
$\left[\because x = \frac{\pi }{4}+h\right]$
$= \displaystyle \lim_{h\to 0} \frac{\left(\frac{1+\tan\,h}{1-\tan\,h}\right)-1}{\cos\left(\frac{\pi }{2}+2h\right)}$
$= \displaystyle\lim_{h\to 0} \frac{1+\tan\,h-1+\tan\,h}{-\sin\,2h\left(1-\tan\,h\right)}$
$= \displaystyle\lim_{h\to 0} \frac{-2\,\tan\,h}{2\,\sin\,h \,\cos\,h\left(1-\tan\,h\right)}$
$= \displaystyle\lim_{h\to 0} \frac{-1}{\cos^{2}\,h\left(1-\tan\,h\right)} = -1$