L'hopitals theorem (Limits)

  • Thread starter ibysaiyan
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  • #1
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Homework Statement



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 [itex]\rightarrow[/itex] pi/2.


Homework Equations





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 [itex]x^{1/x}[/itex]/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
 
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Answers and Replies

  • #2
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You are missing that [itex]\frac{0}{\infty}[/itex] is not a form of indetermination:
[itex]\frac{0}{\infty}=0[/itex]
 
  • #3
SammyS
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0/∞ shouldn't pose a problem.

DiracR beat me !
 
  • #4
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You are missing that [itex]\frac{0}{\infty}[/itex] is not a form of indetermination:
[itex]\frac{0}{\infty}=0[/itex]

0/∞ shouldn't pose a problem.

DiracR beat me !
So I can't solve this problem using the aforementioned theorem ?
 
  • #5
111
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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
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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.

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

[itex]\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}}[/itex][itex]=\frac{0^{\pm}}{+\infty}=0^{\pm}[/itex]

This is the way to solve this limit...
 
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  • #8
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no, with substitution I mean simply to put the value x=pi/2 in the limits.

[itex]\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}[/itex]

This is the way to solve this limit...

Sorry but I am further confused. :(
 
  • #9
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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
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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.

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 ;)
 
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