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- ksaimouli

limts

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- ksaimouli

limts

- katieb

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- ksaimouli

http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__91.gif

- ksaimouli

- anonymous

You can use L'Hopital's rule whereby you take the limit of the derivative of the numerator over the limit of the derivative of the denominator. All with respect to "t".

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- ksaimouli

yes i did that but stuck

- anonymous

Are you able to calculate the derivative of the numerator and the derivative of the denominator? You got stuck? Ok, I'll help a little further.

- ksaimouli

so \[\sec(\frac{ \pi }{ 4 })^2-(\sec \frac{ \pi }{4 })^2\]

- ksaimouli

?

- ksaimouli

do u know what is the answer it is 2

- ksaimouli

but it is 2 i am sure

- anonymous

The numerator goes to:
sec^2 [(1/4)pi] as "t" goes to "0"
So, yes, it goes to
"2"

- anonymous

Yes, definitely, "2" is the answer.

- ksaimouli

how did u get that

- anonymous

The derivative of the numerator is:
sec^2 [(1/4)pi + t]
and that goes to:
sec^2 [(1/4)pi] as "t" goes to "0"
The derivative of the denominator is just "1"
So, you deal with just:
sec^2 [(1/4)pi]
and that = 2

- ksaimouli

what did u do with other half sec^2 [(1/4)pi + t]- sec^2 [(1/4)pi )

- anonymous

The other half is just a constant, so when you take the derivative of that, it disappears. That was where I made an initial mistake but I corrected that.

- ksaimouli

chain rule does not apply for the first one

- anonymous

Here's a graph of the function and you can see that the limit is "2".

- anonymous

The chain actually does apply to the first term, but it doesn't matter, because the derivative of (pi)/4 + t is just "1", so you are multiplying that derivative by "1", so it doesn't become apparent.

- ksaimouli

alright thx

- anonymous

Thx for the recognition btw.

- ksaimouli

for long conversation

- anonymous

And you're welcome. Sorry I went off on a tangent (no pun intended) at the beginning.

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