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

Oct 19, 2014

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

$- 1 \le \sin x \le 1$

$R i g h t a r r o w - 3 \le 3 \sin x \le 3$

$R i g h t a r r o w - \frac{3}{e} ^ x \le \frac{3 \sin x}{e} ^ x \le \frac{3}{e} ^ x$.

Since

${\lim}_{x \to \infty} \left(- \frac{3}{e} ^ x\right) = - \frac{3}{\infty} = 0$

and

${\lim}_{x \to \infty} \frac{3}{e} ^ x = \frac{3}{\infty} = 0$,

we conclude that

${\lim}_{x \to \infty} \frac{3 \sin x}{e} ^ x = 0$

by Squeeze Theorem.

I hope that this was helpful.

Mar 11, 2017

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 $\sin \left(x\right)$ 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 $\sin \left(x\right)$, and the limit is $0$.