# Homework Help: L'hopitals theorem (Limits)

1. Oct 12, 2011

### ibysaiyan

1. The problem statement, all variables and given/known data

The problem is:
A function in the form of f(x)/g(x) is given: [x^2 - pi^2/4 ]/tan^2(x) as x $\rightarrow$ pi/2.

2. Relevant equations

3. The attempt at a solution

Surely I can't solve this question using L'hopital's theorem since it's applicable to indeterminate in the form of 0/0, infinity/infinity, 0 (+- infinity),etc. The above function gives 0/infinity... or am I missing something.. Could I perhaps use trig identity for the denominator ?

Another problem: Also would differentiating $x^{1/x}$/x-1 as x -> 1 give me an answer of e^1 only ?

Here's how I attempted to solve this problem:
I made y = x^1/x which gave me lny = lnx/x ( which's 0/1) that eventually gives me lny = 1 ? :s

Last edited: Oct 12, 2011
2. Oct 12, 2011

### DiracRules

You are missing that $\frac{0}{\infty}$ is not a form of indetermination:
$\frac{0}{\infty}=0$

3. Oct 12, 2011

### SammyS

Staff Emeritus
0/∞ shouldn't pose a problem.

DiracR beat me !

4. Oct 12, 2011

### ibysaiyan

So I can't solve this problem using the aforementioned theorem ?

5. Oct 12, 2011

### DiracRules

There is no reason to evaluate this limit using a theorem. You can solve it by "substitution", and the form you get is not indeterminate.

6. Oct 12, 2011

### ibysaiyan

Thanks for being patient with me but when you say 'substitution' are you referring to trig. identity = sec^2(x) =1+ tan^2 (x) ?
I was handed over this question sheet, where we have to basically evaluate functions using 'hopitals theorem.

7. Oct 12, 2011

### DiracRules

no, with substitution I mean simply to put the value x=pi/2 in the limits.

$\lim_{x\rightarrow\frac{\pi}{2}}\frac{x^2-\frac{\pi^2}{4}}{\tan(x)^2}=\frac{x^2-\frac{\pi^2}{4}}{\tan(x)^2}|_{x=\frac{\pi}{2}}$$=\frac{0^{\pm}}{+\infty}=0^{\pm}$

This is the way to solve this limit...

Last edited: Oct 12, 2011
8. Oct 12, 2011

### ibysaiyan

Sorry but I am further confused. :(

9. Oct 12, 2011

### DiracRules

I edited my previous post, get a glance at it.

The thing is, you don't need any theorem.

The way to solve this limit is the first thing you should have studied, that is: trying substituting the value and see what happens.

10. Oct 12, 2011

### ibysaiyan

Yes, when I sub. in the values I get the following fraction : 0/infinity..
I know if it had been 0/0 or infinity/ infinity.. then I could have used L'hopitals theorem

EDIT: I suppose that's the answer since it's not in the form of 0/0 or infinity/ infinity.

Thanks for your help DiracRules! btw. dirac indeed does ;)

Last edited: Oct 12, 2011