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How do you determine the limit of #e^x/(x-5)^3# as x approaches 5-?

Answer 1

#- oo#

#lim_(x to 5^-) e^x/(x-5)^3#

If we start by simply subbing in #x = 5#, we get:

#e^5/(5-5)^3#. The numerator is finite but the denominator is zero, telling us that the limit is one of either: #pm oo#.

And, because the limit is #x to 5^-#, the denominator is the cube of a negative number, ie it is negative. To convince yourself of that, imagine #x = 4.9999#.

It follows that the limit is #- oo#

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Answer 2

Emily Ammerman

To determine the limit of e^x/(x-5)^3 as x approaches 5-, we can use L'Hôpital's Rule. Taking the derivative of both the numerator and denominator separately, we get (e^x)/(3(x-5)^2). Substituting x = 5 into this expression, we find that the limit is 1/75.

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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.