## ksaimouli Group Title limts one year ago one year ago

1. ksaimouli Group Title
2. ksaimouli Group Title

@tcarroll010

3. tcarroll010 Group Title

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".

4. ksaimouli Group Title

yes i did that but stuck

5. tcarroll010 Group Title

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.

6. ksaimouli Group Title

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

7. ksaimouli Group Title

?

8. ksaimouli Group Title

do u know what is the answer it is 2

9. ksaimouli Group Title

but it is 2 i am sure

10. tcarroll010 Group Title

The numerator goes to: sec^2 [(1/4)pi] as "t" goes to "0" So, yes, it goes to "2"

11. tcarroll010 Group Title

Yes, definitely, "2" is the answer.

12. ksaimouli Group Title

how did u get that

13. tcarroll010 Group Title

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

14. ksaimouli Group Title

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

15. tcarroll010 Group Title

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.

16. ksaimouli Group Title

chain rule does not apply for the first one

17. tcarroll010 Group Title

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

18. tcarroll010 Group Title

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.

19. ksaimouli Group Title

alright thx

20. tcarroll010 Group Title

Thx for the recognition btw.

21. ksaimouli Group Title

for long conversation

22. tcarroll010 Group Title

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