Using the Sandwich Theorem to Solve Limits Involving Trigonometric Functions

[ravenea]

Homework Statement

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

Homework Equations

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.

 

[vela]

What's your question?

 

[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.

 

[SammyS]

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.