Question

I just had a quick question pertaining to L'Hopitol's Rule. How are you able to use it in the equation of :
....

Answer 1

\[\sqrt{x ^{2}+x} -x\]

Answer 2

This is as the limit approaches infinity.

Answer 3

limit d.n.e

Answer 4

L'Hopital deals with these kind of limits:\[\lim_{x \rightarrow A}\frac{ f(x) }{ g(x) }\]
A can be anything, even infinity. Furthermore, \[\lim_{x \rightarrow A}f(x) = \lim_{x \rightarrow A} g(x) = B\]where B can also be anything, as long as it is the same for f and g.
f and g must be differentiable. In that case:\[\lim_{x \rightarrow A}\frac{ f(x) }{ g(x) }=\lim_{x \rightarrow A}\frac{ f'(x) }{ g'(x) }\]Only if the derivatives are simpler (which they often are) does l'H help...
Having said this, I see no use for l'H with your function.

Answer 5

On the other hand, @hbk19 says this limit does not exist.
Try this:\[\lim_{x \rightarrow \infty}\sqrt{x^2+x}-x=\]\[\lim_{x \rightarrow \infty}\left( (\sqrt{x^2+x}-x)*\frac {\sqrt{x^2+x}+x }{ \sqrt{x^2+x}+x } \right)=\]\[\lim_{x \rightarrow \infty} \frac{ x^2+x-x^2 }{\sqrt{x^2+x}+x }=\]\[\lim_{x \rightarrow \infty}\frac{ x }{ \sqrt{x^2+x}+x }\]This last limit looks promising! If you can make a good estimate of the denominator, you'll get a nice limit after all...

If not, then....well l'Hopital's your friend ;)

Answer 6

you can use it in calculating the limits of fractions when both numerator and denominator limit to infinity or to zero.and the approach is that you should derivate nominator and denominator separately