How do you find the limit as x approaches #pi/4# of #[sin(x) - cos(x)] / cos(2x)#?

2 Answers
bp
Apr 16, 2015

#sqrt2/(-2)#

This is an indeterminate form of the type #0/0#, hence L'Hopital's rule would apply and limit can be evaluated by differentiating numerator and denominator and then applying the limit. Accordingly,

Lim #x->pi/4# #(sinx -cosx)/cos(2x)#

= Lim#x->pi/4# #(cosx +sinx)/(-2sin2x)#

= #sqrt2/(-2)#

Jim H
Apr 16, 2015

If you want to evaluate the limit without l'Hopital's Rule (find the limit "algebraically"), do this:

Use
#cos2x = cos^2x-sin^2x = (cosx-sinx)(cosx+sinx)#

#lim_(xrarr pi/4) (sinx-cosx)/cos(2x) = lim_(xrarr pi/4) (- 1(cosx-sinx))/((cosx-sinx)(cosx+sinx))#

#=lim_(xrarr pi/4) (-1)/(cosx+sinx) = (-1)/(1/sqrt2+1/sqrt2)=(-1)/(2/sqrt2) = -sqrt2/2#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
8407 views around the world
You can reuse this answer
Creative Commons License