Question

evaluate limit X tends π/2 (secx-tanx)

Answer 1

lim x --> (pi/2)( sec(x) - tan(x))
lim x-->pi/2 ((1-sin x)/cos x) = lim x-->pi/2 ((1-sin x)(cos x))/ cos^2 x
=lim x-->pi/2 ((1-sin x) cos x)/( 1-sin ^2 x) = lim x-->pi/2 (1-sin x)(cos x)/((1-sin x)(1+sin x))
=lim x-->pi/2 ((cos x)/(1+sin x)) = cos (pi/2)/((1+ sin(pi/2)) = 0/(1+1) = 0

Answer 2

$\displaystyle \sf\lim_{x \to \frac{\pi}{2} } \: (sec \: x - tan \: x) = 0$

Given :

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

To find : The value

Solution :

Step 1 of 2 :

Write down the given limit

The given limit is

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

Step 2 of 2 :

Find the value of the limit

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

$\displaystyle \sf = \lim_{x \to \frac{\pi}{2} } \: \bigg( \frac{1}{cos \: x} - \frac{sin \: x}{cos \: x}\bigg)$

$\displaystyle \sf = \lim_{x \to \frac{\pi}{2} } \: \bigg( \frac{1 - sin \: x}{cos \: x}\bigg) \: \: \: \: \bigg( \: \frac{0}{0} \: \: form\bigg)$

$\displaystyle \sf = \lim_{x \to \frac{\pi}{2} } \: \bigg( \frac{0 - cos \: x}{ - sin \: x}\bigg)$

$\displaystyle \sf = \lim_{x \to \frac{\pi}{2} } \: cot \: x$

$\displaystyle \sf = cot \: \frac{\pi}{2}$

$\displaystyle \sf = 0$

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