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**Prove It** $\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}x^2\sin{\frac{1}{x}} &= \lim_{x \to \infty}x\cdot \frac{\sin{\frac{1}{x}}}{\frac{1}{x}} \end{align*} $

Now if you let $\displaystyle \displaystyle h = \frac{1}{x}$ and note that as $\displaystyle \displaystyle x \to \infty, h \to 0$, then

$\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}x \cdot \frac{\sin{\frac{1}{x}}}{\frac{1}{x}} &= \lim_{h \to 0}\frac{1}{h} \cdot \frac{\sin{h}}{h} \\ &= \lim_{h \to 0}\frac{1}{h} \cdot \lim_{h \to 0}\frac{\sin{h}}{h} \\ &= \infty \cdot 1 \\ &= \infty \end{align*}$