Question

$\huge\boxed{\texttt{fcolorbox{red}{aqua}{Challenge}}}$


$\bold{find\:the\:value\:of}$

$\lim_{x \to \frac{\pi}{2}} ( \sec \: x \: - \: \tan \: x \: )$

Answer 1

I am using L'Hopital's rule for solving this question.

In L'Hopital's rule, we differentiate numerator and denominator both w.r.t. x.

Answer is zero (0).

Answer 2

$\huge\bold{Hay\:mate}}$


$\bold{Your\:answer\:is.\:\:0}$


$\bold{step\:by\:step\:solution →}$

$\lim_{x \to \frac{\pi}{2}} ( \sec \: x \: - \: \tan \: x \: ) \: \: \:$



$\lim _{x \to \frac{\pi}{2} } \: [ \frac{1 - sin \: x}{cos \: x} ]$



$\lim _{x \to \frac{\pi}{2} } \: [ \frac{( {cos}^{2} \frac{1}{2}x \: + \: {sin}^{2} \frac{1}{2} x \: ) - 2 \: sin \frac{1}{2} x \: cos \frac{1}{2}x }{ {cos}^{2} \frac{1}{2}x - {sin}^{2} \frac{1}{2}x } ]$




$\lim _{x \to \frac{\pi}{2} } \: [ \frac{(cos \: \frac{1}{2}x - sin \frac{1}{2} x ) {}^{2} }{(cos \frac{1}{2}x - sin \frac{1}{2}x)(cos \frac{1}{2} x + sin \frac{1}{2}x) } ] \:$




$\lim _{x \to \frac{\pi}{2} } \: \frac{(cos \frac{1}{2} x - sin \frac{1}{2}x) }{(cos \frac{1}{2}x + sin \frac{1}{2} x) }$





$\lim _{x \to \frac{\pi}{2} } \frac{cos \frac{1}{4} \pi - sin \frac{1}{4} \pi}{cos \frac{1}{4}\pi + sin \frac{1}{4} \pi}$






$= \frac{(1/ \sqrt{2}) - (1 / \sqrt{2}) }{(1/ \sqrt{2}) + (1/ \sqrt{2}) } \\ \\ \\ \\ \\ = \frac{0}{2/ \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0$





I hope it's help you




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