MHB 1.6.1 AP Calculus Exam Limits with L'H

[karush]

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

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 ...

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

 

[karush]

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 $\frac{1}{2}(2(1- 0))= 1$.
.