# `lim_(x->0) (e^x - 1 - x)/(x^2)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->0) (e^x - 1 - x)/(x^2)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Expert Answers

Neethu | Certified Educator

Given the limit function `lim_{x->0}(e^x-1-x)/x^2` . We have to find the limit.

We can see that the above limit is of the form `0/0` . So we have to apply L'Hospital's rule. i.e. differentiating both numerator and denominator term we get,

`lim_{x->0}(e^x-1)/(2x)=0/0`

Again it is of the form `0/0` and so we apply the l'hospital rule again to obtain,

`lim_{x->0}e^x/2=e^0/2=1/2`

hence the limit is 1/2.