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ksaimouli
 4 years ago
limts
ksaimouli
 4 years ago
limts

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ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__91.gif

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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".

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0yes i did that but stuck

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are 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
 4 years ago
Best ResponseYou've already chosen the best response.0so \[\sec(\frac{ \pi }{ 4 })^2(\sec \frac{ \pi }{4 })^2\]

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0do u know what is the answer it is 2

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0but it is 2 i am sure

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The numerator goes to: sec^2 [(1/4)pi] as "t" goes to "0" So, yes, it goes to "2"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, definitely, "2" is the answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0what did u do with other half sec^2 [(1/4)pi + t] sec^2 [(1/4)pi )

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0chain rule does not apply for the first one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here's a graph of the function and you can see that the limit is "2".

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thx for the recognition btw.

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0for long conversation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And you're welcome. Sorry I went off on a tangent (no pun intended) at the beginning.
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