# Homework Help: Limit using Sandwich Theorem

1. Aug 21, 2011

### ravenea

1. The problem statement, all variables and given/known data

$\lim_{x \to 0} \sqrt{x^3 + x^2}\sin\frac{\pi}{x}$

2. Relevant equations

3. The attempt at a solution

We know that $-1 \leq \sin\frac{\pi}{x} \leq 1$
$\Leftrightarrow -\sqrt{x^3 + x^2} \leq \sqrt{x^3 + x^2}\sin\frac{\pi}{x} \leq \sqrt{x^3 + x^2}$ since $\sqrt{y} \geq 0\forall y\in\mathbb{P}\cup\{0\}$
Now, $\lim_{x \to 0} -\sqrt{x^3 + x^2} = 0 = \lim_{x \to 0} \sqrt{x^3 + x^2}$
$\therefore \lim_{x \to 0} \sqrt{x^3 + x^2}\sin\frac{\pi}{x} = 0$ by Sandwich Theorem.

Last edited: Aug 21, 2011
2. Aug 21, 2011

### vela

Staff Emeritus

3. Aug 21, 2011

### ravenea

I want to know if it is correct.

What worries me is if I can state that $-1 \leq \sin{\frac{\pi}{x}} \leq 1$ since x is approaching 0, and right there $\frac{\pi}{x}$ is undefined. Thanks in advance.

4. Aug 21, 2011

### SammyS

Staff Emeritus
What you can say is -1 ≤ sin(θ) ≤ 1 for all real θ.

Therefore, $-1 \leq \sin{\frac{\pi}{x}} \leq 1$ provided that x ≠ 0 .

Since x approaches zero for your limit, x is not equal to zero, so you're fine.