Prove a limit exists and evaluate it

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Discussion Overview

The discussion revolves around evaluating the limit of the expression lim x-> 0 (1 - cos(2x))/(x^3 + 4x^2). Participants explore the application of L'Hôpital's rule and the use of derivatives to analyze the limit's existence and value.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant proposes using L'Hôpital's rule after determining that both the numerator and denominator approach 0 as x approaches 0.
  • Another participant asserts that the limit can be evaluated as f''(0)/g''(0) and calculates it to be 4/8, which simplifies to 1/2.
  • Some participants express uncertainty about the implications of f''(0)/g''(0) being 0/0 and whether this indicates that the limit does not exist.
  • A later reply challenges the assertion that the limit will go to infinity if f''(0)/g''(0) equals 0/0, suggesting that L'Hôpital's rule can be applied multiple times in such cases.
  • One participant mentions the potential use of the power series expansion of cosine as an alternative approach.
  • Another participant expresses a sense of progress in understanding the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the limit's existence or value, with multiple competing views on the application of L'Hôpital's rule and the implications of the derivatives evaluated at zero.

Contextual Notes

There are unresolved assumptions regarding the application of L'Hôpital's rule and the conditions under which the limit can be evaluated. The discussion includes differing interpretations of the results of the derivatives.

james.farrow
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We are required to find

lim x-> 0 1- cos(2x)/x^3 + 4x^2

Let f(x) = 1 - cos(2x) & g(x) = x^3 + 4x^2

f'(x) = 2sin(2x) so f'(0)=0
g'(x) = 3x^2 + 8x & g'(0) = 0

so by l'hospital's rule

lim x -> 0 2sin2x/3x^2 + 8x if it exists

Now

f''(x) = 2(2cos2x) & g''(x) = 6x + 8 = 2(3x + 4)

But f''(0) = 4 and g''(0) = 8

Shouldn't f''(0) = g''(0) = 0 here?

I'm a little bit stuck now?

or is the limit just f''(0)/g''(0) which is 4/8 which is a half?

Many thanks

James
 
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The limit is f''(0)/g''(0), and it is 4/8=1/2.

If f"(0)/g''(0)=0/0, then the limit will go to infinity, hence no limit will exist.

Regards.
 
You could use the power series of the cosine.
 
Дьявол said:
If f"(0)/g''(0)=0/0, then the limit will go to infinity, hence no limit will exist.

Regards.

BTW that is simply not true. You can keep applying L'Hoptials rule as many times as you like, while ever you are faced with 0/0. Take for example x^4 / ( 1-cos x )^2, you'd need to go to the 4th derivative in this case before L'Hopitial's finally cracks it.
 
Cheers everyone I think I'm getting the hang of it...
 

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