Limit involving nested roots How do I solve it? Is there some crafty algebraic step I need to follow?
I'm guessing it's -∞ in the end..
$$ \lim_{x\to-∞} \sqrt{x^2-\sqrt{x^2+1}}+x $$

Rewrite it as follows: for $x< 0$,
$$\begin{align}
 \sqrt{x^2-\sqrt{x^2+1}}+x &= \sqrt{x^2-|x|\sqrt{1+\frac{1}{x^2}}}+x = |x|\sqrt{1-\frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}}+x \\
&= -x\left(1-\sqrt{1-\frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}}\right)
\end{align}$$
Writing $u(x)\stackrel{\rm def}{=} \frac{1}{|x|}\sqrt{1+\frac{1}{x^2}}$, we have $u(x)\xrightarrow[x\to-\infty]{} 0$, and by a first-order Taylor expansion
$$
\sqrt{1-u(x)} = 1-\frac{u(x)}{2} + o(u(x))
$$
when $x\to-\infty$. Therefore, we get, when $x\to-\infty$, 
$$\begin{align}
 \sqrt{x^2-\sqrt{x^2+1}}+x
&= x\left(1-\sqrt{1-u(x)}\right)
= x\left( 1-1+\frac{u(x)}{2} + o(u(x)) \right) \\
&= \frac{1}{2}xu(x) + o(xu(x))
\end{align}$$
and since $xu(x)=-\sqrt{1+\frac{1}{x^2}} + o(1) \xrightarrow[x\to-\infty]{} -1$, we eventually obtain
$$
 \sqrt{x^2-\sqrt{x^2+1}}+x
\xrightarrow[x\to-\infty]{} \frac{1}{2}\cdot -1 = -\frac{1}{2}.
$$
