Homework Help: Limits question

1. Nov 16, 2016

Nipuna Weerasekara

1. The problem statement, all variables and given/known data
Solve the following limit.
$$\lim_{x\to0} \space \frac {sin(\pi (Cos ^2 (x)))}{\pi (Cos ^2 (x))}$$

3. The attempt at a solution
When I plug $x\to 0$ in to the limit, I get 0/1... Then what can I do? See here I can't even apply L'Hopital's law... Please help!

Here I see the answer is 0 obviously... But I need to plug this result into another limit question. Which is the real challenge here...
For further guidance,
$$\lim_{x\to0}\space \frac {sin(\pi (Cos ^2 (x)))}{x^2}$$
This is the limit question... The challenge is to get the answer to this limit by without using the L'Hopital's law...

2. Nov 16, 2016

Nipuna Weerasekara

Actually the real question here is to solve the limits question $\lim_{x\to0}\space \frac {sin(\pi (Cos ^2 (x)))}{x^2}$ without using L'Hopital's law...

3. Nov 16, 2016

Nipuna Weerasekara

Just an easy fix... $\;\sin(\pi\cos^2x)=\sin(\pi-\pi\sin^2x)=\sin(\pi\sin^2x)$

4. Nov 16, 2016

Nipuna Weerasekara

The answer is $\pi$

5. Nov 16, 2016

Ray Vickson

The form 0/1 is perfectly OK; it just gives you 0. The only thing that is forbidden is dividing by 0; dividing 0 by something else nonzero is absolutely allowed, and always gives 0.

If your original question was not the one you were interested in, why did you pose it?

Anyway, if you are not allowed to use l'Hospital's rule, what ARE you allowed to use?

Note added in edit: posts #3 and #4 did not appear on my screen until after I submitted this response.

6. Nov 16, 2016

Nipuna Weerasekara

This is not a case whether you are allowed to use the L'Hopital law or not... It is a challenge where you cannot use it. Hence the question gets tricky...
The following shows the method to solve the limit without using the law.
$\lim_{x\to0}\space \frac {sin(\pi (Cos ^2 (x)))}{x^2}$
$\;\sin(\pi\cos^2x)=\sin(\pi-\pi\sin^2x)=\sin(\pi\sin^2x)$
$\lim_{x\to0}\space \frac {\sin(\pi\sin^2x)}{x^2}$
$\lim_{x\to0}\space \frac {\sin(\pi\sin^2x)}{x^2 (\pi\sin^2x)}{(\pi\sin^2x)}$
$x \to 0 \space$ then $\space \pi \sin^2x \to 0$
$\lim_{\pi \sin^2x \to 0} \space \frac {Sin(\pi \sin^2x)}{\pi \sin^2x} \space \lim_{x\to 0}\space \frac {\pi\sin^2x}{x^2}$
$(1) \space. \space \lim_{x\to0}\space \frac {\pi\sin^2x}{x^2}$
$\pi \space \lim_{x\to0}\space \frac {\sin^2x}{x^2}$
$\pi \space .\space (1)$
$\pi$