# Limit of square root function.

1. Oct 9, 2012

### 11thHeaven

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.

2. Oct 9, 2012

### clamtrox

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

3. Oct 9, 2012

### 11thHeaven

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 :)

4. Oct 9, 2012

### HallsofIvy

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

5. Oct 9, 2012

### mtayab1994

Multiply and divide by √(x^2+x)+x

6. Oct 10, 2012

### 11thHeaven

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.

7. Oct 10, 2012

### ehild

Divide both the numerator and denominator by x.

ehild

8. Oct 10, 2012

### 11thHeaven

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

9. Oct 10, 2012

### 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).

10. Oct 10, 2012

### micromass

Staff Emeritus