Limit of square root without L'Hopital's rule. How might one go about taking the following limit without using L'Hopital's rule? I am stumped:
$$\lim_{x \to \infty} \sqrt{x^2 + x} - x$$
 A: $$\sqrt{x^2+x}-x=\sqrt{x^2\left(1+\frac1x\right)}-x=x\sqrt{1+\frac1x}-x=x\left(\sqrt{1+\frac1x}-1\right)\;;$$
let $u=\dfrac1x$, and this becomes $$\frac{\sqrt{1+u}-1}u=\frac{\sqrt{1+u}-1}u\cdot\frac{\sqrt{1+u}+1}{\sqrt{1+u}+1}=\frac{u}{u\left(\sqrt{1+u}+1\right)}=\frac1{\sqrt{1+u}+1}\;.$$
As $x\to\infty$, $u\to0^+$, so just calculate $$\lim_{u\to0^+}\frac1{\sqrt{1+u}+1}\;.$$
A: Hint: 
Multiply and divide function under limit by $ \sqrt{x^2 + x} + x$: $\lim\limits_{x \to \infty}\left( \sqrt{x^2 + x} - x\right)=\lim\limits_{x \to \infty}\dfrac{\left( \sqrt{x^2 + x} - x\right)\left( \sqrt{x^2 + x} + x\right)}{ \sqrt{x^2 + x} + x}$
A: $$\lim_{x\to\infty} \sqrt{x^2+x}-x=\lim_{x\to\infty} (\sqrt{x^2+x}-x)\cdot \frac{ \sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty} \frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\dots$$
Can you take it from here?
A: HINT
$$\sqrt{x^2+x} - x = \dfrac{\left(\sqrt{x^2+x} - x\right) \left(\sqrt{x^2+x} + x \right)}{\sqrt{x^2+x} + x} = \dfrac{x^2+x - x^2}{\sqrt{x^2+x} + x}$$
A: Observing that for $x>0$ we have $$\sqrt{x^2+x}-x=|x|\sqrt{1+\frac1x}-x=x\left(\sqrt{1+\frac1x}-1\right),$$ then using $(a-b)(a+b)=a^2-b^2$ gives us $$\left(\sqrt{x^2+x}-x\right)\cdot\left(\sqrt{1+\frac1x}+1\right)=x\left(1+\frac1x-1\right)=1.$$ Thus, $$\sqrt{x^2+x}-x=\frac1{\sqrt{1+\frac1x}+1},$$ which makes evaluation of the limit as $x\to\infty$ much simpler.
A: You could complete the square inside the square root to give:
$$\sqrt{\left(x+\frac 1 2\right)^2-\frac 1 4} - x=\left(x+\frac 1 2\right)\sqrt{1-\frac 1 {(2x+1)^2}}-x$$
And use the binomial series expansion of the square root. I mention this because it is a different idea from that in the other answers - it is not necessarily a better one. But if you are taking a factor outside the square root, I think completing the square gives you a better chance of getting the right factor. This puts a square in the denominator, rather than a linear denominator as in some of the other suggestions.
A: Hint: make a fraction of it and use $(a+b)(a-b)$.
A: Second hint: rewrite the fraction you've made (per Berci's comment) into $\frac{1}{\sqrt{1+\frac{1}{x}}+1}.$
A: First, factor out $x$ from the radical:
$$\lim_{x \to \infty} x \left(\sqrt{1+\frac{1}{x}} - 1 \right)$$
You can then Taylor expand the radical, as $1/x$ is small as $x$ becomes large.  The relevant expansion is $(1+1/x)^{1/2} \approx 1 + \frac{1}{2x}$.  This should make the result of the limit apparent.
