PBM.1 Limit to Zero: $$\lim_{x\to 0} \frac{\cos 3x-1}{x^2}$$

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Discussion Overview

The discussion revolves around evaluating the limit $$\lim_{x\to 0} \frac{\cos 3x-1}{x^2}$$. Participants explore different methods for solving this limit, including the use of L'Hôpital's Rule and algebraic manipulation.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant applies L'Hôpital's Rule and arrives at a limit of $$-\frac{9}{2}$$, questioning the correctness of their approach.
  • Another participant confirms the correctness of the first participant's result and suggests a slightly different application of L'Hôpital's Rule to reach the same conclusion.
  • A third participant expresses agreement with the previous contributions, affirming the correctness of the limit evaluation.
  • A fourth participant proposes a substitution method, transforming the limit into a different variable, which also leads to the same result of $$-\frac{9}{2}$$.
  • Some participants express skepticism about the necessity of the substitution, suggesting that the limit can be evaluated without it, while acknowledging that it may appear cleaner with the substitution.

Areas of Agreement / Disagreement

Participants generally agree on the final result of the limit being $$-\frac{9}{2}$$, but there is some disagreement regarding the necessity and clarity of using substitution in the evaluation process.

Contextual Notes

The discussion includes various approaches to the limit, with some participants favoring L'Hôpital's Rule while others prefer algebraic manipulation. The effectiveness and clarity of each method are debated, but no consensus is reached on the necessity of substitution.

karush
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$$\lim_{{x}\to{0}}\frac{\cos\left({3x}\right)-1}{{x}^{2}}$$
$$\frac{f'}{g'}
=-\frac{3\sin\left({3x}\right)}{2x}
=-\frac{9}{2}\cdot\frac{\sin\left({3x}\right)}{3x}$$
$x\to 0$ is $-\frac{9}{2}$

Just seeing if this is correct or better way to do it
 
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That's correct, but if you are allowing yourself L'Hôpital, then I would just write:

$$L=\lim_{x\to0}\frac{\cos(3x)-1}{x^2}=-\frac{3}{2}\lim_{x\to0}\frac{\sin(3x)}{x}=-\frac{9}{2}\lim_{x\to0}\cos(3x)=-\frac{9}{2}$$
 
Correct! Well done!
 
$$\lim_{x\to0}\dfrac{\cos3x-1}{x^2}$$

$$x=\dfrac13y$$

$$\lim_{y\to0}\dfrac{9(\cos y-1)}{y^2}$$

$$\lim_{y\to0}\dfrac{-9\sin^2y}{y^2(\cos y+1)}=-9\cdot1\cdot\dfrac12=-\dfrac92$$
 
Sorry but I don't see the purpose of sticking $y$ in this thing
 
karush said:
Sorry but I don't see the purpose of sticking $y$ in this thing

Yes, you could do it without making a substitution, but it just looks cleaner if you do. However, the main takeaway from Greg's post is how to take the limit without using L'Hôpital's Rule at all. :)
 

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