Determining Limits Using the Squeeze Theorem Practice Test

Since $$\sin\left(\frac{1}{x}\right)$$ oscillates, the limit does not exist by the Squeeze Theorem

Since $$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$, we have $$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$

Since $$0 \leq \sin\left(\frac{1}{x}\right) \leq 1$$, we have $$0 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$

The Squeeze Theorem cannot be applied because $$\sin\left(\frac{1}{x}\right)$$ is undefined at $$x = 0$$