What is equal ? lim_(x->pi/2)sin(cosx)/(cos^2(x/2) -sin^2(x/2) )= ?

3 Answers
MattyMatty
Mar 5, 2018

1

Explanation:

"Note that : "color(red)(cos^2(x)-sin^2(x) = cos(2x))
"So here we have"

lim_{x->pi/2} sin(cos(x))/cos(x)

"Now apply rule de l' Hôptial :"

= lim_{x->pi/2} cos(cos(x))*(-sin(x))/(-sin(x))
= lim_{x->pi/2} cos(cos(x))
= cos(cos(pi/2))
= cos(0)
= 1

Ratnaker Mehta
Mar 5, 2018

1.

Explanation:

Here is a way to find the limit without using L'Hospital's Rule :

We will use, lim_(alpha to 0)sinalpha/alpha=1.

If we take cosx=theta, then as x to pi/2, theta to 0.

Replacing cos^2(x/2)-sin^2(x/2) by cosx=theta, we have,

:."The reqd. lim."=lim_(theta to 0)sintheta/theta=1.

maganbhai P.
Mar 5, 2018

1

Explanation:

We know that,
color(red)(cosA=cos^2(A/2)-sin^2(A/2))
So,
L=lim_(x->pi/2)(sin(cosx))/(cos^2(x/2)-sin^2(x/2))=lim_(x->pi/2)(sin(cosx))/(cosx)
Take,cosx=theta,
We get, xto(pi/2)rArrtheta tocos(pi/2)rArrtheta to0.
:.L=lim_(theta->0)(sintheta)/theta=1

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