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..
$$ \lim_{x\to-∞} \sqrt{x^2-\sqrt{x^2+1}}+x $$
 A: Let $-1/x=h\implies h\to0^+$
$\sqrt{1+x^2}=\dfrac{\sqrt{1+h^2}}h$
$x^2-\sqrt{1+x^2}=\dfrac{1-h\sqrt{1+h^2}}{h^2}$
$\implies\sqrt{x^2-\sqrt{1+x^2}}=\dfrac{\sqrt{1-h\sqrt{1+h^2}}}h$
$$\lim_{x\to-\infty} \sqrt{x^2-\sqrt{x^2+1}}+x =\lim_{h\to0^+}\dfrac{\sqrt{1-h\sqrt{1+h^2}}-1}h$$
Now rationalize the numerator by multiplying the  numerator  & the denominator by  $\sqrt{1-h\sqrt{1+h^2}}+1$
A: $$\lim_{x\to-\infty}\sqrt{x^2-\sqrt{x^2+1}}+x=\lim_{x\to-\infty}\left(\sqrt{x^2-\sqrt{x^2+1}}+x\right)\cdot \frac{\sqrt{x^2-\sqrt{x^2+1}}-x}{\sqrt{x^2-\sqrt{x^2+1}}-x}=$$
$$=\lim_{x\to-\infty}\frac{-\sqrt{x^2+1}}{\sqrt{x^2-\sqrt{x^2+1}}-x}=\lim_{x\to-\infty}\frac{(-\sqrt{x^2+1})/x}{(\sqrt{x^2-\sqrt{x^2+1}}-x)/x}$$
$$=\lim_{x\to-\infty}\frac{-\sqrt{1+1/x^2}}{\sqrt{1-\sqrt{1/x^2+1/x^4}}-1}=-\infty$$
A: $=\lim_{t\to \infty} \sqrt{t^2 -\sqrt{t^2 +1}} -t =\lim_{t\to \infty} \frac{-\sqrt{t^2 +1}}{\sqrt{t^2 -\sqrt{t^2 +1}} +t}=\lim_{t\to \infty} \frac{-\sqrt{1 +\frac{1}{t^2}}}{\sqrt{1 -\sqrt{\frac{1}{t^2} +\frac{1}{t^4}}} +1}=-\frac{1}{2}$
A: 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
$$
 \sqrt{x^2-\sqrt{x^2+1}}+x
\xrightarrow[x\to-\infty]{} \frac{1}{2}\cdot -1 = -\frac{1}{2}.
$$
