ksaimouli
  • ksaimouli
limts
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ksaimouli
  • ksaimouli
http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__91.gif
ksaimouli
  • ksaimouli
@tcarroll010
anonymous
  • 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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More answers

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

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