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.