Simplifying the Limit: $\frac{\cos x - \sqrt{1 + \sin^2 x}}{x^2}$

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Homework Help Overview

The discussion revolves around evaluating the limit \(\lim_{x \to 0}\frac{\cos x - \sqrt{1 + \sin^2 x}}{x^2}\), which involves trigonometric functions and their behavior as \(x\) approaches zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for simplifying the limit, including the use of double angle formulas, multiplying by the conjugate, and applying the Binomial theorem. There is also a question about the applicability of L'Hôpital's rule in this context.

Discussion Status

There is a divergence in opinions regarding the limit's value, with some participants asserting it is \(-1\) while others reference an answer of \(1/2\). The discussion includes attempts to clarify the methods used and whether L'Hôpital's rule is permissible, indicating an ongoing exploration of the problem.

Contextual Notes

Participants note that L'Hôpital's rule is not allowed for this particular test, which raises questions about alternative approaches to limits that may be in indeterminate form.

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Homework Statement


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


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?
 
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I don't think the limit is 1/2.

Are you allowed to use L'Hopital's rule?
 
kru_ said:
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.
 
-Dragoon- said:

Homework Statement


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


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 \displaystyle \cos(x) + \sqrt{1 + \sin^{2}(x)}

The numerator then becomes -2sin2(x)
 
Or alternatively employ cos x ~ (1 - 0.5x^2) and sin x ~ x, and Binomial theorem to the first order on the numerator.
 
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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