Question

How do you find the limit of `e^(-5x)` as x approaches `oo`?

Answer 1

0

Explanation:

`e^(ax)=1+(ax)+(ax)^2/2!+(ax)^3/3!+...to oo`, as `xtooo`, when a > 0.

So, th reciprocal `e^(-ax) to 0`, as `x to oo`.

Here, a = 5 > 0.

Answer 2

The limit is `0`. Here is an informal approach.

Explanation:

As `x` increases without bound (as `x` gets bigger and bigger),

the product `5x` also gets bigger and bigger,

so, the exponent `-5x` is a negative number farther and farther from `0` (more and more negative) ( big negative number)

Wait, `e^(-5x) = 1/e^(5x)`.

Now `x rarroo`, so `5x rarroo` and `e^(5x) rarroo`.

So we have `1` over a number that is going to infinity, so the ratio is going to `0`.