anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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

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anonymous
  • anonymous
you can keep going so long as you are in indeterminate form and the derivatives exist
anonymous
  • anonymous
Thanks a lot ;)

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