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Calculus1 15 Online
OpenStudy (anonymous):

evaluate limit ( sqrt(n^2+n) - n )

OpenStudy (p0sitr0n):

assuming we are taking the limit to infinity, simply multiply the top and the bottom by the conjugate of the numerator, i.e. \[(\sqrt{n^2+n}+n)\] Then this will allow you to simplify your fraction up to \[\lim_{n \rightarrow \infty}\frac{n}{\sqrt{n^2+n}+n}\] Now factor and simplify

OpenStudy (anonymous):

I have reached to as limit n goes to infinity (n/sqrt(n^2+n) +n)

OpenStudy (anonymous):

Dividing by n ?

OpenStudy (p0sitr0n):

ok good. Now what you can do is factor out the n^2 out of the square root, which will give you \[n(\sqrt{1+\frac{1}{n}}+1)\] in the bottom

OpenStudy (p0sitr0n):

yeah, then you cancel both n, and are left with an expression that you can evaluate

OpenStudy (anonymous):

Canceling from above ?

OpenStudy (anonymous):

Yes I get it now

OpenStudy (anonymous):

Thank you

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