Limit Question: a better way to solve this limit? where $x \to -\infty$ Hey guys so I have this limit:
$$\lim_{x \to -∞} f(x) = {(x+\sqrt{x^2+2x})}$$
I solved it by multiplying numerator and denominator by $$x-\sqrt{x^2+2x}$$ and got $-1$ as my answer, but I really don't like how I solved it; any better way to solve it?
 A: You are looking, setting $t=|x|$, at
$$
\lim_{t\to\infty} \sqrt{t^2-2t} - t
$$
For the sake of it, here is a solution using the Taylor expansion of $\sqrt{1+u}$ around $0$: for $t>0$,
$$
\sqrt{t^2-2t} - t
= t\sqrt{1-\frac{2}{t}} - t
= t\left(\sqrt{1-\frac{2}{t}} - 1\right)
= t\left(1-\frac{1}{t} + o\left(\frac{1}{t}\right) - 1\right)
= -1+o(1)
$$
showing the limit is indeed $-1$.
A: You can set $x=-\dfrac1t\enspace(t>0,\,t\to 0)$. The expression rewrites as
$$f(x)=-\frac1t+\sqrt{\frac{1\mathstrut{}}{t^2}-\frac2t}=\frac{-1+\sqrt{\strut 1-2t}}t.$$
This is the rate of variation of the function $\sqrt{1-2t}\;$ from $t=0$, so the limit is
$$\bigl(\sqrt{1-2t}\bigr)'_{t=0}=\frac1{2\sqrt{1-2t}}\cdot(-2)\Biggm|_{t=0}=-1.$$
A: For fun
$t=-x$, let $t >2.$
$(t^2-2t+1-1)^{1/2}-t=$
$((t-1)^2-1)^{1/2} -((t-1)^2)^{1/2}-1;$
Then with $m:=(t-1)^2$.
$(m-1)^{1/2}- m^{1/2} -1$.
$y_m:=(m-1)^{1/2}-m^{1/2}.$
$\lim_{m \rightarrow \infty} y_m=0$.
Can you prove it?
A: I suggest to flip the sign by $y=-x\to \infty$ then
$$\lim_{x \to -∞} f(x) = {x+\sqrt{x^2+2x}}=\lim_{y \to \infty} f(x) = {-y+\sqrt{y^2-2y}}$$
and then
$$\sqrt{y^2-2y}-y=\left(\sqrt{y^2-2y}-y\right)\cdot\frac{\sqrt{y^2-2y}+y}{\sqrt{y^2-2y}+y}$$
Edit As another alternative we can use binomial series $t\to 0, \,(1+t)^n=1+nt+O(t^2)$, are you aware about that?
We proceed as follows
$$\sqrt{y^2-2y}=\sqrt{y^2}\cdot \sqrt{1-2/y}=y\cdot(1-2/y)^\frac12=y\cdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$
then
$$\sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) \to -1$$
Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then
$$|O(1/y)|\le \frac c y \to 0 \implies O(1/y)\to 0$$
