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What is the process involved in finding the value of this limit?
Lim x -> infinity
square root(x^2+x)-x
Please explain the steps, more then the answer, I really need the solution. Thank you! :)
2 Answers
- Anonymous9 years agoFavorite Answer
If you know L'Hopital's rule, one way is to manipulate to get the form 0/0 or infinity/infinity:
sqrt{x^2 +x} - x
= x*[sqrt{1 + 1/x} - 1]
= [sqrt{1 + 1/x} - 1]/(1/x)
THe limit as x ---> infinity gives 0/0, so use L'Hopital:
derivative of numerator = (1/2)(1/sqrt{1+1/x})*(-1/x^2)
derivative of denominator = -1/x^2
(derive of num)/(derive of den) = (1/2)1/sqrt(1+1/x) --->1/2
- Anonymous4 years ago
Use polar coordinates to start this (x = r cos t, y = r sin t): lim((x,y)?(0,0)) (x^2 + y^2) / [?(x^2+y^2+25) - 5] = lim(r?0+) r^2 / [?(r^2 + 25) - 5] Now rationalize the denominator by conjugates: lim(r?0+) r^2 [?(r^2 + 25) + 5] / {[?(r^2 + 25) - 5][?(r^2 + 25) + 5]} = lim(r?0+) r^2 [?(r^2 + 25) + 5] / [(r^2 + 25) - 25] = lim(r?0+) [?(r^2 + 25) + 5] = 5 + 5 = 10. i'm hoping this helps!