Question

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

Answer 1

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