Limit of square root function.

[11thHeaven]

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

 

[clamtrox]

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

 

[11thHeaven]

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]

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

 

[mtayab1994]

Multiply and divide by √(x^2+x)+x

 

[11thHeaven]

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]

Divide both the numerator and denominator by x.

ehild

 

[11thHeaven]

Divide both the numerator and denominator by x.

ehild

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

 

[11thHeaven]

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).

 

[micromass]

Please start a new thread for each new question.