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

1 Answer
Eddie
Feb 27, 2017

#- oo#

Explanation:

#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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