Evaluating $\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$. Why doesn't $x^2$ overpower $2x$ when $x\to\infty$, so that the answer is $0$ (instead of $1$)? 
Evaluate $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$$

Hello!
I was solving this problem, and here is my approach:

Inside the square root $x^2$ overpowers $2x$ as $x \to \infty$, so $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x = \lim_{x \to \infty} \sqrt{x^2} - x = \lim_{x \to \infty} |x| - x = 0$$ which is wrong answer. Actual answer is $1$.

This is the way I have solved so many problems when $x \to \infty$, especially when the terms are in denominator.
For example, consider $$\lim_{x \to \infty} \frac{x^2 + x}{2x^2 + 4x + 10}$$
Now, we ignore the smaller power terms, and answer is $\frac{1}{2}$. Why can we ignore smaller powers here but not in the original question on top?
Sorry if this is a stupid question.

$\lim_{x \to \infty} \sqrt{x^2+2x}-x =$
$\lim_{x \to \infty} (\sqrt{x^2+2x}-x) \cdot \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{(\sqrt{x^2+2x}-x)\cdot(\sqrt{x^2+2x}+x)}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{|x^2+2x|-x^2}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{x^2+2x-x^2}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{2x}{\sqrt{x^2+2x}+x} =$
$\lim_{x \to \infty} \frac{2}{\frac{\sqrt{x^2+2x}}{x}+1} =$
$\lim_{x \to \infty} \frac{2}{\sqrt{1+\frac{2}{x}}+1} =$
$\frac{2}{\sqrt{1+\lim_{x \to \infty} \frac{2}{x}}+1} = 1$
