Math, asked by sahil214, 1 year ago

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

Answers

Answered by simmusebastion
176
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
Attachments:
Answered by pulakmath007
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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