# Power Series and Limits

## Key Questions

• 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.

• Here is a simple application of a power series in evaluating a limit.

${\lim}_{x \to 0} \frac{\sin x}{x}$

by replacing $\sin x$ by its Maclaurin series.

=lim_{x to 0}{x-x^3/{3!}+x^5/{5!}-x^7/{7!}+cdots}/{x}

by distributing the division to each term,

=lim_{x to 0}(1-x^2/{3!}+x^4/{5!}-x^6/{7!}+cdots)

by sending $x$ to zero,

$= 1 - 0 + 0 - 0 + \cdots$

since all but the first term are zero,

$= 1$

I hope that this was helpful.