How do I evaluate this limit?

lim x*sin(1/x)

x->infinity

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- Sep 16th 2011, 01:03 PMSneakyhelp with limits
How do I evaluate this limit?

lim x*sin(1/x)

x->infinity - Sep 16th 2011, 01:09 PMSironRe: help with limits
Rewrite:

$\displaystyle \lim_{x\to \infty} \sin\left(\frac{1}{x}\right)\cdot x=\lim_{x\to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$

You can use the substitution, let $\displaystyle \frac{1}{x}=u$. If $\displaystyle x\to \infty$ then $\displaystyle u \to ?$, afterwards you should recognize a standard limit. - Sep 16th 2011, 01:10 PMPlatoRe: help with limits
- Sep 16th 2011, 01:16 PMSneakyRe: help with limits
sin(h) / h as x-> 0 should be 0, but how do I formally show that? And why are we considering x->0 when we need x->infinity?

edit: after its sin(h)/h, can I use L'hopital rule and derive the top and bottom to get 1*cos(h) / 1, then then sub h as 1/x, then sub x as infinity, which results to 0, and cos of 0 is 1, so the limit is 1? - Sep 16th 2011, 01:35 PMPlatoRe: help with limits
- Sep 16th 2011, 01:56 PMmr fantasticRe: help with limits
- Sep 16th 2011, 02:03 PMSneakyRe: help with limits
Okay, I now understand why we get

lim h-> 0 sin(h)/h, but I'm still not sure why the limit of this is 0, if you plug in 0 in h, then you get 0/0. - Sep 16th 2011, 02:06 PMmr fantasticRe: help with limits
- Sep 16th 2011, 08:57 PMProve ItRe: help with limits
Or better yet, do a little research.

Think of the unit circle.

http://i22.photobucket.com/albums/b3...itcircle-1.jpg

If the radius is one unit in length and angle made from the radius and the horizontal axis (measured in radians) is $\displaystyle \displaystyle \theta$, then the green length is $\displaystyle \displaystyle \sin{\theta}$, the red length is $\displaystyle \displaystyle \cos{\theta}$, and the purple length is $\displaystyle \displaystyle \tan{\theta}$.

Clearly, the area of the smallest triangle is a little less than the area of the circular sector, which is a little less than the area of the larger triangle. So

$\displaystyle \displaystyle \begin{align*} \frac{1}{2}\sin{\theta}\cos{\theta} \leq \frac{1}{2}\theta &\leq \frac{1}{2}\tan{\theta} \\ \sin{\theta}\cos{\theta} \leq \theta &\leq \tan{\theta} \\ \sin{\theta}\cos{\theta} \leq \theta &\leq \frac{\sin{\theta}}{\cos{\theta}} \\ \cos{\theta} \leq \frac{\theta}{\sin{\theta}} &\leq \frac{1}{\cos{\theta}} \\ \frac{1}{\cos{\theta}} \geq \frac{\sin{\theta}}{\theta} &\geq \cos{\theta} \\ \cos{\theta} \leq \frac{\sin{\theta}}{\theta} &\leq \frac{1}{\cos{\theta}} \\ \lim_{\theta \to 0}\cos{\theta} \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} &\leq \lim_{\theta \to 0}\frac{1}{\cos{\theta}} \\ 1 \leq \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} &\leq 1 \end{align*}$

It should be clear therefore, that $\displaystyle \displaystyle \lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} = 1$.

(Actually, we have really only proven the right-hand limit, but the proof of the left-hand limit is almost identical.)