- #1

karush

Gold Member

MHB

- 3,267

- 4

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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- MHB
- Thread starter karush
- Start date

- #1

karush

Gold Member

MHB

- 3,267

- 4

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

- #2

skeeter

- 1,104

- 1

you should be familiar with this basic limit ...

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

- #3

karush

Gold Member

MHB

- 3,267

- 4

i saw another example of the problem but expanded into confusion

that was slam dunk

mahalo

- #4

HOI

- 923

- 2

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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