Question

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

Answer 1

`- 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`