Recall the bounds of the sine function.

Question

ScanSolve AI Tutor · Accepted Answer

Here's how to solve the limit problem using the Squeeze Theorem:

Recall the bounds of the sine function.
We know that the sine function is always bounded between $-1$ and 1, regardless of its argument.
$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$
Multiply all parts of the inequality by $x^2$ .
Since $x^2 \geq 0$ for all real $x$ , multiplying by $x^2$ does not change the direction of the inequalities.
$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$
Take the limit as $x \to 0$ for all three parts of the inequality.
We evaluate the limits of the bounding functions:
$\lim_{x \to 0} (-x^2) = -(0)^2 = 0$ $\lim_{x \to 0} (x^2) = (0)^2 = 0$
Apply the Squeeze Theorem.
Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$ , and our function $x^2 \sin\left(\frac{1}{x}\right)$ is "sandwiched" between them, the Squeeze Theorem states that the limit of the middle function must also be 0.
$0 \leq \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \leq 0$
Therefore,
$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = \boxed{\text{0}}$

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