How do you compute the limit of #(sinx-cosx)/cos(2x)# as #x->pi/4#?

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Answer 1 · 2016-11-19T05:17:51

Multiply by #frac{sin(x) + cos(x)}{sin(x) + cos(x)}#.

#frac{sin(x) - cos(x)}{cos(2x)} * frac{sin(x) + cos(x)}{sin(x) + cos(x)} = frac{sin^2(x) - cos^2(x)}{cos(2x)(sin(x) + cos(x))}#

Use the double angle identity for cosine: #cos^2(x) - sin^2(x) -= cos(2x)#

#frac{sin^2(x) - cos^2(x)}{cos(2x)(sin(x) + cos(x))} = frac{-cos(2x)}{cos(2x)(sin(x) + cos(x))}#

Now to evaluate the limit

#lim_{x->pi/4}frac{sin(x) - cos(x)}{cos(2x)} = lim_{x->pi/4}-frac{1}{sin(x) + cos(x)}#

#= -frac{1}{sin(pi/4) + cos(pi/4)}#

#= -frac{1}{1/sqrt2 + 1/sqrt2}#

#= sqrt2/2#