1. L'Hopital

Let $\displaystyle f(x)=x^2\sin{1/x}$ for $\displaystyle x\neq 0$ , let $\displaystyle g(x)=\sin{x}$ for $\displaystyle x\in\mathbb{R}$ .
Show that $\displaystyle \lim_{x\to 0}f(x)/g(x) = 0$ but that $\displaystyle \lim_{x \to 0}f'(x)/g'(x)$ does not exist.

2. Originally Posted by Kipster1203
Let $\displaystyle f(x)=x^2\sin{1/x}$ for $\displaystyle x\neq 0$ , let $\displaystyle g(x)=\sin{x}$ for $\displaystyle x\in\mathbb{R}$ .
Show that $\displaystyle \lim_{x\to 0}f(x)/g(x) = 0$ but that $\displaystyle \lim_{x \to 0}f'(x)/g'(x)$ does not exist.
You are like the seventh person to post this exact question, and I will give the same answer the seventh time. There is no subtlety to this. Just do it.

what's the issue?

3. Lol if I could just "do it" I wouldn't be wasting my time writing this.

I took the first derivative, and have the following:
$\displaystyle \frac{-2\cos{\frac{1}{x}}}{x\cos{x}}$

Now I took it two more times but it doesn't get me anywhere, I know there's something I'm not seeing, some help would be really nice...

4. The derivative of f(x) is 2x sin(1/x)- cos(1/x) and the derivative of g(x) is cos(x). $\displaystyle \frac{f'}{g'}= \frac{2x sin(1/x)- cos(1/x)}{cos(x)}$, not what you give. It should be obvious that the limit of that fraction does not exist.

5. Don't tell people stuff is obvious when you have no idea what level they're at, if we're asking a question it's obviously cause we don't know the answer. We can't all be MHF Experts like you guys, sorry.

6. Originally Posted by Kipster1203
Don't tell people stuff is obvious when you have no idea what level they're at, if we're asking a question it's obviously cause we don't know the answer. We can't all be MHF Experts like you guys, sorry.
Ok, so assuming that HallsOfIvy's derivative is correct (which there is no need to assume) then you want to see why $\displaystyle \lim_{x\to0}\frac{2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)}{\cos(x)}$ doesn't exist. So, assume that it does. Then, it equals $\displaystyle L$ and so $\displaystyle L=1\cdot L=\lim_{x\to0}\cos(x)\cdot L=\lim_{x\to0}\cos(x)\cdot\lim_{x\to0}\frac{2x\sin (\frac{1}{x})-\cos(\frac{1}{x})}{\cos(x)}=$$\displaystyle \lim_{x\to0}\left(2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)\right)$. I need to assume that the limit existed so that I could combine the limits in the last part and clear out that cosine. So, by assuming that our limit existed we've show that $\displaystyle \lim_{x\to0}\left(2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)\right)=\lim_{x\to0}f( x)$ exists as well. But, look at $\displaystyle f(a_n),f(b_n)$ where $\displaystyle a_n=\frac{1}{2\pi n}$ and $\displaystyle b_n=\frac{1}{\frac{\pi}{2}n}$

7. lol

If you want to see why it is "obvious"...

if $\displaystyle x\sim 0$, $\displaystyle \cos x \sim 1$. So the denominator is $\displaystyle \sim 1$...

$\displaystyle {2x\sin(\frac1x)-\cos(\frac1x) \over \cos x} \sim {2x\sin(\frac1x)-\cos(\frac1x) \over 1} = 2x\sin(\frac1x)-\cos(\frac1x)$

... and since sine is bounded, if $\displaystyle x\sim 0$ then $\displaystyle 2x\sin(\frac1x) \sim 2\cdot0\cdot \sin(\frac1x)=0$ so...

$\displaystyle 2x\sin(\frac1x)-\cos(\frac1x) \sim -\cos(\frac1x)$

...and this oscillates infinitly as $\displaystyle x\to 0$ just like $\displaystyle -\cos(y)$ oscillates as $\displaystyle y\to\pm\infty$

:]

(oh dont hurt me, I can be rigorous I swear!)

8. Originally Posted by Kipster1203
Don't tell people stuff is obvious when you have no idea what level they're at, if we're asking a question it's obviously cause we don't know the answer. We can't all be MHF Experts like you guys, sorry.
"Have no idea what level they're at"? This was a problem involving derivatives and limits. Aren't we allowed to assume they know that? And if so, then it is obvious that this limit does not exist.