Re: limit as x -> infinity?

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**Barthayn** How does one show that the limit as x approaches ∞ $\displaystyle x^2 sin \frac {1}{x}$ is equal to ∞ mathematically? My solution book claims it is just the limit as h approaches zero from the right $\displaystyle \frac {1}{h} \frac {sin h}{h}$ is equal to ∞. How does the solution booklet workout? I don't see the logic in it. (Shake)

$\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

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

Re: limit as x -> infinity?

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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

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

thanks I see it now.