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http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__91.gif
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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yes i did that but stuck
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.
so \[\sec(\frac{ \pi }{ 4 })^2-(\sec \frac{ \pi }{4 })^2\]
?
do u know what is the answer it is 2
but it is 2 i am sure
The numerator goes to: sec^2 [(1/4)pi] as "t" goes to "0" So, yes, it goes to "2"
Yes, definitely, "2" is the answer.
how did u get that
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
what did u do with other half sec^2 [(1/4)pi + t]- sec^2 [(1/4)pi )
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.
chain rule does not apply for the first one
Here's a graph of the function and you can see that the limit is "2".
1 Attachment
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.
alright thx
Thx for the recognition btw.
for long conversation
And you're welcome. Sorry I went off on a tangent (no pun intended) at the beginning.

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