## ksaimouli 2 years ago limts

1. ksaimouli
2. ksaimouli

@tcarroll010

3. tcarroll010

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

yes i did that but stuck

5. tcarroll010

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

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

7. ksaimouli

?

8. ksaimouli

do u know what is the answer it is 2

9. ksaimouli

but it is 2 i am sure

10. tcarroll010

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

11. tcarroll010

Yes, definitely, "2" is the answer.

12. ksaimouli

how did u get that

13. tcarroll010

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

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

15. tcarroll010

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

chain rule does not apply for the first one

17. tcarroll010

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

18. tcarroll010

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

alright thx

20. tcarroll010

Thx for the recognition btw.

21. ksaimouli

for long conversation

22. tcarroll010

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