1. ## These crazy limits..

I understand how to work with basic limits, and those ones under square roots, & how to implement that sinx/x=1 identity (& cos, tan).

But now, I'm coming across some really confusing stuff & my teacher simply is incapable of expressing these in intelligible terms.

here are some examples:
$
\displaystyle \lim_{x \to 0} \frac{ x^2 }{ \cos 8 x - \cos 4 x } =
$

$
\displaystyle \lim_{x \to 1^{+}} ( \displaystyle \frac {1}{\ln x} - \displaystyle \frac {1}{x - 1} ) =
$

$
\displaystyle \lim_{x \to + \infty} \left( 1 + \frac {3}{x} \right)^{x/2} =
$

$
\displaystyle \lim_{x\to \pi^+} \left(4x-4\pi \right)\tan\!\left(\frac{x}{2}\right)
$

I'm not asking for answers to all of these. But can some one point me to some videos/material that can teach me how to solve & understand these particular types of limits? I have no idea where to begin!

Perhaps a sample calculation for just the first one too.

Thanks

2. I'll give you a heads up for the first one. This is the url. Good luck. These problems tend to be long. But you can do it.

http://i52.tinypic.com/4pz8zp.jpg

3. Thanks!

Is L'Hopital's rule needed for all of these?

4. Originally Posted by Vamz
Thanks!

Is L'Hopital's rule needed for all of these?
Well, can they be put into a form where l'Hopspital's Rule can be used?

The third one can be done simply by making a simple substitution and using a well known limit definition for e^x ....

Moderator edit: The third question was double posted (against forum rules, by the way) and has been further discussed here: http://www.mathhelpforum.com/math-he...tml#post582776.

5. Originally Posted by Vamz
I understand how to work with basic limits, and those ones under square roots, & how to implement that sinx/x=1 identity (& cos, tan).

But now, I'm coming across some really confusing stuff & my teacher simply is incapable of expressing these in intelligible terms.

here are some examples:
$
\displaystyle \lim_{x \to 0} \frac{ x^2 }{ \cos 8 x - \cos 4 x } =
$

$
\displaystyle \lim_{x \to 1^{+}} ( \displaystyle \frac {1}{\ln x} - \displaystyle \frac {1}{x - 1} ) =
$

$
\displaystyle \lim_{x \to + \infty} \left( 1 + \frac {3}{x} \right)^{x/2} =
$

$
\displaystyle \lim_{x\to \pi^+} \left(4x-4\pi \right)\tan\!\left(\frac{x}{2}\right)
$

I'm not asking for answers to all of these. But can some one point me to some videos/material that can teach me how to solve & understand these particular types of limits? I have no idea where to begin!

Perhaps a sample calculation for just the first one too.

Thanks
If you're interested, for my senior High School project I wrote a 100 page manual on how to compute difficult limits, I could send it to you. Besides that I don't know many places where you can find material on "techniques" of limit evaluation.

Here's a thread you might find interesting from my past life

6. ## Limit regarding pi

$
\displaystyle \lim_{x\to \pi^+} \left(4x-4\pi \right)\tan\!\left(\frac{x}{2}\right)
$

$
\displaystyle let f(x)=(4x-4\pi) ; g(x)=\tan(\frac{x}{2})
$

so, therefore
$
\displaystyle Lim( f(x)* g(x) ) = lim F(x) * lim G(x)
$

for g(x)
$
\displaystyle U=\frac{x}{2}
$

$
\displaystyle U*\frac{\tan U}{U} = U * 1 = U
$

So, the limit for g(x) is $\displaystyle\frac{\pi}{2}$

Now, for dealing with f(x)
$
\displaystyle (4x-4\pi) =0
$

no matter how I move these variables around, f(x) always ends up zero, sending my entire limit to zero - which is incorrect! What am I missing here?

Thanks!

7. Originally Posted by Vamz
$
\displaystyle \lim_{x\to \pi^+} \left(4x-4\pi \right)\tan\!\left(\frac{x}{2}\right)
$

$
\displaystyle let f(x)=(4x-4\pi) ; g(x)=\tan(\frac{x}{2})
$

so, therefore
$
\displaystyle Lim( f(x)* g(x) ) = Lim f(x) * Lim g(x)
$

for g(x)
$
\displaystyle U=\frac{x}{2}
$

$
\displaystyle U*\frac{\tan U}{U} = U * 1 = U
$

So, the limit for g(x) is $\displaystyle\frac{\pi}{2}$

Now, for dealing with f(x)
$
\displaystyle (4x-4\pi) =0
$

no matter how I move these variables around, f(x) always ends up zero, sending my entire limit to zero - which is incorrect! What am I missing here?

Thanks!
$\lim\limits_{x\to\pi^+}\tan\left(\frac{x}{2}\right )=-\infty$. But $\lim\limits_{x\to\pi^+}(4x-4\pi)\tan\left(\frac{x}{2}\right)\to 0\cdot -\infty\neq 0$!!!

Thus, rewrite the limit as follows:

$\lim\limits_{x\to\pi^+}\dfrac{4x-4\pi}{\cot\left(\frac{x}{2}\right)}\rightarrow \dfrac{0}{0}$

At this stage, now apply L'Hopitals rule. Can you proceed?

8. Agh, thats awesome. Thanks.