Question

I have two questions .. let's suppose we want to solve a limit by using L'Hopital's Rule , and we can't solve it because there's still infinity ... can we use more than once derivatives.... Example:
http://i1064.photobucket.com/albums/u368/Kreshnik-joker/Limit.jpg

(I didn't try if this limit works or not...)

2. question nr 2. is about vectors, since I'm dumb about these kind of problems, I could use a little help here :) .

http://i1064.photobucket.com/albums/u368/Kreshnik-joker/Vectors.jpg

Thank you.

Answer 1

For question 1, it is difficult to find out to find the answer at the first sight as it has a sqrt
To cancel the sqrt, multiply its conjugate. Don't forget to eliminate the conjugate using the denominator.
(sqrt(x^2 +1) - x)(sqrt(x^2 +1) + x) / (sqrt(x^2 +1) + x)
for numerator, it is in the form of (a+b)(a-b). By expanding it, the sqrt can be cancel and we can get 1.
we can now get 1/(sqrt(x^2 +1) + x).
By direct substitution, we can get 1/infinity which is =0
Here's the answer! :D

Answer 2

multiply by the conjugate for the first one
\[(\sqrt{x^2+1}-x)(\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x})=\frac{x^2+1-x}{\sqrt{x^2+1}+x}\]
\[=\frac{1}{\sqrt{x^2+1}+x}\] now take the limit as x goes to infinity and get 0

Answer 3

have read my question?? ... @satellite73 & @CoCoTsoi ... thanks a lot, I've solved this one... but i want to know: "can we use derivatives more than once, if we want to solve it by L'Hospital's Rule ?" ... :D

Answer 4

you can keep going so long as you are in indeterminate form and the derivatives exist

Answer 5

Thanks a lot ;)