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How do you evaluate #(e^x - 1 - x )/ x^2# as x approaches 0?

Answer 1

1/2

#lim_(x->0) (e^x -1 -x)/x^2 = (e^0 -1 -0)/0 = (1-1)/0 = 0/0#

This is an indeterminate type so use l'Hopital's Rule. That is, take the derivative of the top and the bottom and then find the limit of its quotient.

#lim_(x->0) (e^x-1)/(2x) =( e^0 -1)/0 = (1-1)/0 = 0/0# This is still an indeterminate form so let's use l'Hopital's Rule again.

#lim_(x->0) e^x /2 = e^0 /2 = 1/2#

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Answer 2

Elijah Abram

To evaluate the expression (e^x - 1 - x) / x^2 as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (e^x - 1 - 1) / (2x). Evaluating this expression as x approaches 0, we have (1 - 1 - 1) / (2 * 0) = -1/0, which is undefined.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.