Solve Limit: $$\lim_{x\to0} \frac{sin(\pi(Cos^2(x)))}{\pi(Cos^2(x))}$$

[Nipuna Weerasekara]

Homework Statement

Solve the following limit.
$$\lim_{x\to0} \space \frac {sin(\pi (Cos ^2 (x)))}{\pi (Cos ^2 (x))}$$

The Attempt at a Solution

When I plug $x\to 0$ into 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...

 

[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...

 

[Nipuna Weerasekara]

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

 

[Nipuna Weerasekara]

The answer is $\pi$

 

[Ray Vickson]

Homework Statement

Solve the following limit.
$$\lim_{x\to0} \space \frac {sin(\pi (Cos ^2 (x)))}{\pi (Cos ^2 (x))}$$

The Attempt at a Solution

When I plug $x\to 0$ into 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...

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.

 

[Nipuna Weerasekara]

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.

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 $\space \pi \sin^2x \to 0$
$\lim_{\pi \sin^2x \to 0} \space \frac {Sin(\pi \sin^2x)}{\pi \sin^2x} \space \lim_{x\to 0}\space \frac {\pi\sin^2x}{x^2}$
$(1) \space. \space \lim_{x\to0}\space \frac {\pi\sin^2x}{x^2}$
$\pi \space \lim_{x\to0}\space \frac {\sin^2x}{x^2}$
$\pi \space .\space (1)$
$\pi$