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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