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

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

Calculus1
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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
I have reached to as limit n goes to infinity (n/sqrt(n^2+n) +n)
Dividing by n ?

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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
yeah, then you cancel both n, and are left with an expression that you can evaluate
Canceling from above ?
Yes I get it now
Thank you

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