Question

limts

Answer 1

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

Answer 2

@tcarroll010

Answer 3

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".

Answer 4

yes i did that but stuck

Answer 5

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.

Answer 6

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

Answer 7

?

Answer 8

do u know what is the answer it is 2

Answer 9

but it is 2 i am sure

Answer 10

The numerator goes to:

sec^2 [(1/4)pi] as "t" goes to "0"

So, yes, it goes to

"2"

Answer 11

Yes, definitely, "2" is the answer.

Answer 12

how did u get that

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

chain rule does not apply for the first one

Answer 17

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

Answer 18

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.

Answer 19

alright thx

Answer 20

Thx for the recognition btw.

Answer 21

for long conversation

Answer 22

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