help with limits

**Sneaky** · September 16th 2011, 01:03 PM

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

**Siron** · September 16th 2011, 01:09 PM

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

You can use the substitution, let $\frac{1}{x}=u$ . If $x\to \infty$ then $u \to ?$ , afterwards you should recognize a standard limit.

**Plato** · September 16th 2011, 01:10 PM

Originally Posted by Sneaky

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

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

$\lim _{h \to 0} \frac{{\sin (h)}}{h} = ?$

**Sneaky** · September 16th 2011, 01:16 PM

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?

**Plato** · September 16th 2011, 01:35 PM

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.

$\lim _{h \to 0} \frac{{\sin (h)}}{h}$ is the same as
$\lim _{x \to \infty } x\sin \left( {\frac{1}{x}} \right)$ if $h = \frac{1}{x}$ .

**mr fantastic** · September 16th 2011, 01:56 PM

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).

**Sneaky** · September 16th 2011, 02:03 PM

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.

**mr fantastic** · September 16th 2011, 02:06 PM

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.

**Prove It** · September 16th 2011, 08:57 PM

Or better yet, do a little research.

Think of the unit circle.

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

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

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

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

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

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