Question

How do I find `lim_(x->oo)(3sin(x))/e^x` using power series?

Answer 1

To be honest, I would not use power series on this one since this is a perfect problem to demonstrate the application of Squeeze Theorem. Here is how:

We know

`-1 le sinx le 1`

`Rightarrow -3 le 3sinx le 3`

`Rightarrow -3/e^x le {3sinx}/e^x le 3/e^x`.

Since

`lim_{x to infty}(-3/e^x)=-3/infty=0`

and

`lim_{x to infty}3/e^x=3/infty=0`,

we conclude that

`lim_{x to infty}{3sinx}/e^x=0`

by Squeeze Theorem.

---

I hope that this was helpful.

Answer 2

The power series in question are:

• `sin(x)=sum_(n=0)^oo(-1)^nx^(2n+1)/((2n+1)!)=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...`

• `e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...`

So:

`lim_(xrarroo)(3sin(x))/e^x=(3(x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...))/(1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...)`

In the numerator, we see that `sin(x)` alternates and skips even powers of `x`. On the other hand, `e^x` is always increasing and skips no powers, so it is evident that `e^x` will rise much faster than `sin(x)`, and the limit is `0`.