ksaimouli
  • ksaimouli
limts
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.