# help with limits

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• Sep 16th 2011, 01:03 PM
Sneaky
help with limits
How do I evaluate this limit?
lim x*sin(1/x)
x->infinity
• Sep 16th 2011, 01:09 PM
Siron
Re: 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 PM
Plato
Re: help with limits
Quote:

Originally Posted by Sneaky
How do I evaluate this limit?
lim x*sin(1/x)
x->infinity

Just note that $\displaystyle x\sin \left( {\frac{1}{x}} \right) = \frac{{\sin \left( {\frac{1}{x}} \right)}}{{\frac{1}{x}}}$.

$\displaystyle \lim _{h \to 0} \frac{{\sin (h)}}{h} = ?$
• Sep 16th 2011, 01:16 PM
Sneaky
Re: 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 PM
Plato
Re: help with limits
Quote:

Originally Posted by Sneaky
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?

Do not use L'hopital rule on these problems.

$\displaystyle \lim _{h \to 0} \frac{{\sin (h)}}{h}$ is the same as
$\displaystyle \lim _{x \to \infty } x\sin \left( {\frac{1}{x}} \right)$ if $\displaystyle h = \frac{1}{x}$.
• Sep 16th 2011, 01:56 PM
mr fantastic
Re: help with limits
Quote:

Originally Posted by Sneaky
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?

sin(h) / h as h-> 0 is a standard limit. You are expected to review your class notes and textbook as part of the process in solving questions like the one you posted (and certainly after getting help with the question).
• Sep 16th 2011, 02:03 PM
Sneaky
Re: 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 PM
mr fantastic
Re: help with limits
Quote:

Originally Posted by Sneaky
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.

The limit is NOT 0. As I have already said, go to your class notes or textbook - review the standard limits you have undoubtedly been given and are expected to know and use.
• Sep 16th 2011, 08:57 PM
Prove It
Re: 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.)