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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

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

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,678

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

- Mar 1, 2012

- 655

you should be familiar with this basic limit ...

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

- Thread starter
- #3

- Jan 31, 2012

- 2,678

i saw another example of the problem but expanded into confusion

that was slam dunk

mahalo

- Jan 30, 2018

- 426

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

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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