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.

Okay.... I'd say the issue of "overpowering" is that it doesn't work when there isn't something else acting an the thing doing the  overpowering to something finite, so that the thing being overpowered is forced to be insignificant.  If the thing doing the overpowering goes to infinity and is only countered by something else going to negative infinity so they cancel out, then the thing being overpowered isn't being forced to go to $0$-- it's merely going ... somewhere else.  Just not as fast as the overpowering thing.
As $x\to \infty$ then the $\sqrt{x^2} \to \infty$ and a higher rate than $\sqrt{+2x} \to \infty$ but that doesn't mean it goes to $0$.
We have $\sqrt{x^2 + 2x}> \sqrt {x^2} = x$ so we must have $\sqrt{x^2+2x}-x = f(x) > 0$.
Now we do know that $f(x) << x$
and as $x\to SomethingREALLYhuge$
we will have $f(x)\to SomethingNOWHEREnearashuge$,
but there is nothing forcing it to zero.  We have no idea at this point what $\lim_{x\to\infty}\sqrt{x^2 + 2x} -x = \lim_{x\to \infty} f(x)$ ought to be.
[Compare this to $\frac {x^5 + x^3}{2x^5 + 4x^2}$ where the $x^5$s "overpower" then $x^3$ and $4x^2$ and breezily say $\lim \frac {x^5 + x^3}{2x^5 + 4x^2}\approx \lim \frac {x^5}{2x^5} = \frac 12$.  In this case we are dividing the $x^5$ by $x^5$ to "overpower to something finite".  So as $x^5$ is being forced to something finite the $x^3$ and $x^2$ are forced to $\frac {finite}{x^2}$ and $\frac {finite}{x^3}$ and go to zero.]
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Now to actually do this thing.  $\sqrt{x^2+2x} = \sqrt{x^2 + 2x + 1 -1}= \sqrt{(x+1)^1 - 1}$.
Now we can say that as $x\to \infty$ that $(x+1)^2$ "overpowers" the $-1$.  The difference between $x^2$ overpowering $-1$ to insignificance and $x^2$ overpowering $2x$ but not to insignificance, is that $-1$ is finite and gets insignificant when compared to very large $x$, whereas $2x$ is not finite.
$\lim \sqrt{x^2 + 2x} -x =\lim \sqrt{(x+1)^2-1} - x\approx \lim \sqrt{(x+1)^2} - x=\lim (x+1) -x = 1$.
.....
Still, informality is only for estimating and intuiting answers.  To actually answer the question we must be more formal and proper.
