1. ## Infinite limits question

The question is
Show that:
$\lim{x \rightarrow\infty}(\surd(x^2+x)-x)=1/2$

I have no idea how I should start. The only possible solution i can come up with is to input different values for x and show that as they get higher the limit gets closer to 1/2...

Is there a better way to do it? Thanks!!

2. Hello,
Originally Posted by natashabu
The question is
Show that:
$\lim{x \rightarrow\infty}(\surd(x^2+x)-x)=1/2$

I have no idea how I should start. The only possible solution i can come up with is to input different values for x and show that as they get higher the limit gets closer to 1/2...

Is there a better way to do it? Thanks!!
Yes, there is a better way

$\lim_{x \to \infty} \sqrt{x^2+x}-x=\lim_{x \to \infty} ~ x \left(\sqrt{1+\frac 1x}-1\right)$

Multiply top and bottom by the conjugate $\sqrt{1+\frac 1x}+1$ :

$\lim_{x \to \infty} ~ x \left(\sqrt{1+\frac 1x}-1\right)=\lim_{x \to \infty} ~ x \cdot \frac{(\sqrt{1+\tfrac 1x}-1)(\sqrt{1+\tfrac 1x}+1)}{\sqrt{1+\tfrac 1x}+1}$

$=\lim_{x \to \infty} ~ x \cdot \frac{\tfrac 1x}{\sqrt{1+\tfrac 1x}+1}=\lim_{x \to \infty} ~ \frac{1}{\sqrt{1+\tfrac 1x}+1}$

Note that $\lim_{x \to \infty} ~ \frac 1x=0$ and find the value of the limit.

3. I get it! Thanks a lot! I wouldn't have been able to figure that out.