How do you find the limit of #(e^x - cos x -2x)/(x^2 -2x) # as x approaches 0?

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How do you find the limit of #(e^x - cos x -2x)/(x^2 -2x) # as x approaches 0?

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Answer 1 · 2016-03-19T23:43:33

Hence the limit #lim_(x->0) [(e^x - cos x -2x)/(x^2 -2x)]=0/0 # yields an undefined form of #0/0# we can apply L'Hôpital's rule to find the limit . So we have that

#lim_(x->0) [(e^x - cos x -2x)/(x^2 -2x)]= lim_(x->0) [[d(e^x - cos x -2x)/dx]/[d(x^2 -2x)/dx)]= lim_(x->0) [e^x+sinx-2]/[2x-2]=1/2#

Finally #lim_(x->0) [(e^x - cos x -2x)/(x^2 -2x)]=1/2#

Refer to L Hopital's rule

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How do you find the limit of #(e^x - cos x -2x)/(x^2 -2x) # as x approaches 0?

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