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Trig limit.

  • Thread starter -Dragoon-
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  • #1
309
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Homework Statement


Find the value of [tex]\lim_{x \to 0}\frac{cosx - \sqrt{1 + sin^{2}x}}{x^{2}}[/tex]


Homework Equations


N/A


The Attempt at a Solution


The answer is 1/2, but I don't know how they got that. I've tried using double angle formulas and multiplying by the conjugate but I get nowhere. How should I attempt to simplify this limit?
 

Answers and Replies

  • #2
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I don't think the limit is 1/2.

Are you allowed to use L'Hopital's rule?
 
  • #3
309
7
I don't think the limit is 1/2.

Are you allowed to use L'Hopital's rule?
It's a past test question and for this test it wasn't allowed to be used, but even then, L'hopital's rule doesn't give 1/2 which is the answer on answer sheet.
 
  • #4
SammyS
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Homework Statement


Find the value of [tex]\lim_{x \to 0}\frac{cosx - \sqrt{1 + sin^{2}x}}{x^{2}}[/tex]


Homework Equations


N/A


The Attempt at a Solution


The answer is 1/2, but I don't know how they got that. I've tried using double angle formulas and multiplying by the conjugate but I get nowhere. How should I attempt to simplify this limit?
That limit is -1 .

To do it without L'Hôpital's rule, multiply the numerator & denominator by [itex]\displaystyle \cos(x) + \sqrt{1 + \sin^{2}(x)}[/itex]

The numerator then becomes -2sin2(x)
 
  • #5
Curious3141
Homework Helper
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Or alternatively employ cos x ~ (1 - 0.5x^2) and sin x ~ x, and Binomial theorem to the first order on the numerator.
 
  • #6
309
7
The answer threw me off, but I used the same method you did Sammy.

Just one last question: Are there limits that would be in indeterminate form but cannot be simplified by using L'hopital's rule? If yes, could you give an example and how would one proceed to determine whether L'hopital's rule works for a simplifying a certain limit or not?
 

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