# 1.6.1 AP Calculus Exam Limits with L'H

• MHB
Gold Member
MHB
$\displaystyle\lim_{x \to 0}\dfrac{1-\cos^2(2x)}{(2x)^2}=$

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

skeeter
$1-\cos^2 (2x) = \sin^2(2x)$

you should be familiar with this basic limit ...

$\displaystyle \lim_{u \to 0} \dfrac{\sin(u)}{u} = 1$

Gold Member
MHB
been i while,,,

i saw another example of the problem but expanded into confusion

that was slam dunk

mahalo

HOI
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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