1.6.1 AP Calculus Exam Limits with L'H

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Discussion Overview

The discussion revolves around evaluating the limit $\displaystyle\lim_{x \to 0}\dfrac{1-\cos^2(2x)}{(2x)^2}$ using L'Hôpital's rule and exploring alternative approaches. The scope includes mathematical reasoning and limit evaluation techniques relevant to calculus.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant notes that the limit results in an indeterminate form $\dfrac{0}{0}$, suggesting the use of L'Hôpital's rule.
  • Another participant points out that $1 - \cos^2(2x)$ can be rewritten as $\sin^2(2x)$ and references the basic limit $\displaystyle \lim_{u \to 0} \dfrac{\sin(u)}{u} = 1$.
  • A participant expresses confusion over a similar problem but acknowledges that this limit seems straightforward.
  • One participant provides a detailed application of L'Hôpital's rule, calculating derivatives and showing the limit approaches 1, but does not assert this as the only method.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the preferred method for solving the limit, with some favoring L'Hôpital's rule while others suggest using the sine limit directly. The discussion remains unresolved regarding the best approach.

Contextual Notes

Some participants rely on specific limit properties and derivatives, which may depend on familiarity with calculus concepts. The discussion does not clarify all assumptions or steps in the reasoning process.

karush
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$\displaystyle\lim_{x \to 0}\dfrac{1-\cos^2(2x)}{(2x)^2}=$

by quick observation it is seen that this will go to $\dfrac{0}{0)}$
so L'H rule becomes the tool to use
but first steps were illusive
the calculator returned 1 for the Limit
 
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$1-\cos^2 (2x) = \sin^2(2x)$

you should be familiar with this basic limit ...

$\displaystyle \lim_{u \to 0} \dfrac{\sin(u)}{u} = 1$
 
been i while,,,

i saw another example of the problem but expanded into confusion

that was slam dunk

mahalo
 
If you really want to do it the "hard way", L'Hopital's rule does work.
The derivative of $1- cos^2(2x)$ is $4 sin(2x)cos(2x)$ and the derivative of $(2x)^2= 4x^2$ is $8x$ so we look at the limit as x goes to 0 of $\frac{4 sin(2x)cos(2x)}{8x}= \frac{1}{2}\frac{sin(2x)cos(2x)}{x}$.
The numerator and denominator of that still go to 0 so use L'Hopital's rule again!
The derivative of $sin(2x)cos(2x)$ is $2cos^2(2x)- 2sin^2(2x)$ and the derivative of x is 1 so now we look at

$\frac{1}{2}\frac{2(cos^2(x)- sin^2(x)}{1}$. At x= 0, cos(x)= 1 and sin(x)= 0 so that is $\frac{1}{2}(2(1- 0))= 1$.
.
 

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