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How do you find the limit of #(e^x - 1)/x^3# as x approaches 0?

Answer 1

Charles Anctil

The limit does not exist, because, as #xrarr0# the function increases without bound. #lim_(xrarr0)(e^x-1)/x^3 = oo#

For his one, I would use l'Hospital's rule.

The initial forms of #lim_(xrarr0)(e^x-1)/x^3# is #0/0#.

Applying l'Hospital, gets us to

#lim_(xrarr0)e^x/(3x^2)# which has the form #1/0#

Because both the numerator and denominator are positive for #x# near #0#, this form tells us that the function increases without bound ad #xrarr0# from either side.

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Answer 2

Samantha Adkison

Infinity.

Differentiate numerator and denominator and take the limit for the ratio applying L' Hospital's rule.for limits of indeterminate forms. The ratio of the derivatives is #e^x/(3x^2)#. #e^0# = 1

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Answer 3

Ezra Adams

To find the limit of (e^x - 1)/x^3 as x approaches 0, we can use L'Hôpital's rule. Taking the derivative of the numerator and denominator separately, we get (e^x) in the numerator and (3x^2) in the denominator. Evaluating the limit of these derivatives as x approaches 0, we find that the limit is 1/6. Therefore, the limit of (e^x - 1)/x^3 as x approaches 0 is 1/6.

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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.