Study of the first and second derivative of $\sqrt{|x^2+x|}-x$ I am not able to study the positivity  of the first and second derivatives of $\sqrt{|x^2+x|}-x$ (that is, the values of $x$ for which the derivatives are positive, negative, or zero), because the expressions turn out to be too complicated for me. However, I must be mistaken, because this exercise is supposed to be easy. Can you show me in detail what steps should I do to solve this problem? 
Also, how do you find the limit of $\sqrt{|x^2+x|}-x$ for $x \to \pm \infty$?
 A: First you're going to want to pay attention to the behavior of $|x^2+x|$. Notice that $x^2+x \geq 0$ for all $x \in (\infty,-1] \cup [0,\infty)$ and $x^2+x<0$ when $x \in (-1,0)$. This means you can make two cases by using properties of the absolute value function.
Case 1:  $x \in (\infty,-1] \cup [0,\infty)$. Then $|x^2+x| = x^2+x$ and hence our derivative pops out easily enough with the chain rule and product rule. You should get $$\left(\sqrt{|x^2+x|}-x\right)' = ((x^2+x)^{1/2}-x)' = \frac{2x+1}{2(x^2+x)^{1/2}}-1$$ and it follows that $$\left(\sqrt{|x^2+x|}-x\right)'' = \frac{1}{(x^2+x)^{1/2}}-\frac{(2x+1)^2}{4(x^2+x)^{3/2}}$$ These derivatives aren't super easy, but they shouldn't be impossible for someone in a calculus class. This should give you a head start on finding when the first and second derivatives are positive in case 1.
Case 2:  $x \in (-1,0)$. Then $|x^2+x| = -x^x-x$. Hence, you need to evaluate $$\left(\sqrt{|x^2+x|}-x\right)' = 
((-x^2-x)^{1/2}-x)'$$ as well as the second derivative. I will leave it to you to solve these derivatives. They should be similar to the ones in case 1.
As for the limit, we are interested in very large magnitudes of $x$. That is, we don't care about $x \in (-1,0)$. So, it suffices to consider $\lim_{x\to \pm \infty} \sqrt{x^2+x}-x$. I'm not sure how formal of a write-up you need, but hopefully this logic can help you see the answer. For large $x$ we know $x^2>>x$. The user LutzL contributed the following logic: $$\lim_{x\to \infty}\sqrt{x^2+x}-x=\lim_{x\to \infty}\sqrt{(x+1/2)^2-1/4}-x\approx \lim_{x\to -\infty}(x+1/2)-x = \frac{1}{2}$$
For large, negative $x$, we know $-x$ is positive and $\sqrt{x^2+x}$ is positive. Thus,
$$\lim_{x\to -\infty} \sqrt{x^2+x}-x \approx  \lim_{x\to -\infty} \sqrt{x^2}-x = \lim_{x \to -\infty} -2x = \infty $$ 
A: As always were differences of square roots are involved, you should check the application of the binomial formulas.
$$
\sqrt{x^2+x}-x=\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2+x}+x}
$$
This readily gives the limit for $x\to+\infty$.
Transforming for $x>0$ as 
$$
\frac1{\sqrt{1+\frac1x}+1}
$$
also allows to reason about monotonicity, since the denomimator is positive and monotonically and strictly decreasing.
