limit as x -> infinity?

**Barthayn** · November 28th 2011, 04:38 PM

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

**Prove It** · November 28th 2011, 04:59 PM

Originally Posted by Barthayn

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

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

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

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

**Barthayn** · November 28th 2011, 05:22 PM

Originally Posted by Prove It

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

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

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

thanks I see it now.

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