# Limit of square root function.

I have to find the limit as x→∞ of √(x2+x)-x

I can't rearrange this into a form where I can put infinity into the expression and get a meaningful answer. I've tried taking out square roots to get √x( √(x+1)-√x ) but if I put infinity into this I just get ∞(∞-∞) which is meaningless.

Now I know that the limit approaches 0.5 (simply by plugging large numbers into my calculator) but I have no idea why this happens.

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Remember the formula (x+y)(x-y) = x2 - y2? That might prove useful here.

Remember the formula (x+y)(x-y) = x2 - y2? That might prove useful here.
Are you suggesting $(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)$? If so, how do I account for the $(\sqrt{x^2+x}+x)$ factor?

Thanks for helping :)

HallsofIvy
Homework Helper
Have you considered "multiplying and dividing by the same thing"?

Multiply and divide by √(x^2+x)+x

Multiply and divide by √(x^2+x)+x
If I do that, then I end up with $\dfrac{x}{\sqrt{x^2+x}+x}$, and I'm not sure how to progress from here. I can use L'Hopital's rule but then I get a horrible mess which doesn't seem to lead to any sort of answer.

ehild
Homework Helper
Divide both the numerator and denominator by x.

ehild

Divide both the numerator and denominator by x.

ehild
Ahhh. That makes the whole thing work, thanks :)

Also, could anyone explain how the limit of $(1+\frac{1}{x})^x=e$ as x goes to infinity? I first assumed it went to 1 (as 1/x goes to 0, and (1+0)^∞=1).