Find all the asymptote of $1-x+\sqrt{2+2x+x^2}$ I find 
$$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=2,$$ but i'am stuck when $x\rightarrow-\infty$ how to find that $y=-2x$ is an oblique asymptote. Any idea? 

General Method
Consider a curve given by the equation $f(x, y) = 0$, where $f(x, y)$ is a polynomial of degree $n$ (in $x$ and $y$). If it contains the term $y^n$ (with some non-zero coefficient), then it has no asymptote parallel to the $y$-axis. Otherwise, equate the total coefficient of the highest degree term of $y$ to 0 and solve (if possible). This gives all asymptotes parallel to the $y$-axis.

Example
$x^2 - xy - y + 1 = 0$
  The degree of the equation is $2$, and $y^2$ is not present, which means the curve might have an asymptote parallel to the $y$-axis. The highest degree term of $y$ is $y$, and its total coefficient is $-(x + 1)$ ("total" here meaning we include not only constant coefficients, but also any $x$-term that is multiplied with $y$). Equating this to zero, $-(x + 1) = 0 \Rightarrow x = -1$ is an asymptote parallel to the $y$-axis.

Similarly, we can find the asymptotes parallel to the $x$-axis as well, but that is unnecessary as the procedure for finding the "oblique" asymptotes does this job as well. Suppose there are asymptotes of the form $y = mx + c$. Substitute $y = mx + c$ in $f(x, y)$ to get a polynomial in $x$. Equate the coefficients of the highest and next highest degree terms of $x$ to zero. This gives two equations in $m$ and $c$ (in general). If these can be solved simultaneously, each solution pair $(m, c)$ corresponds to an asymptote $y = mx + c$.

Example
$x^2 - xy - y + 1 = 0$
  Substituting $y = mx + c$:
  $x^2 - x(mx + c) - (mx + c) + 1 = 0$
Now, we do not have to expand the whole LHS (though it's simple enough in this case). It suffices to find the coefficients of the highest and next highest degree terms. The highest degree term is $x^2$ with coefficient $(1 - m)$. So $1 - m = 0 \Rightarrow m = 1$. The next highest degree term is $x$, with coefficient $-c - m$, and equating this to zero, with $m = 1$, we get $-c - 1 = 0 \Rightarrow c = -1$. Thus, $y = x - 1$ is the other asymptote.

Application
Applying this to the problem at hand, first we rewrite the equation in polynomial form: 
$y = 1 - x + \sqrt{2 + 2x + x^2} \Rightarrow\\
x + y - 1 = \sqrt{2 + 2x + x^2} \Rightarrow$
$(x + y - 1)^2 = 2 + 2x + x^2$.
The degree of is $2$, and there is a $y^2$ term, so there is no asymptote parallel to the $y$-axis. Now, let $y = mx + c$ be an asymptote (not parallel to the $y$-axis). Substituting $y = mx + c$ in the equation:
$(x + mx + c - 1)^2 = 2 + 2x + x^2$
Without expanding the whole LHS, we can see that the highest degree term will be $x^2$ with coefficient $(m + 1)^2$, and on the RHS, $x^2$ has coefficient $1$. So $(m + 1)^2 = 1 \Rightarrow m + 1 = \pm 1 \Rightarrow m = 0, -2$.
The next highest degree term is $x$, with coefficient $2(m + 1)(c - 1)$ on the LHS, and $2$ on the RHS, so $2(m + 1)(c - 1) = 2 \Rightarrow (m + 1)(c - 1) = 1$.
For $m = 0$, $c - 1 = 1 \Rightarrow c = 2$, and for $m = -2$, $1 - c = 1 \Rightarrow c = 0$.
Thus the two asymptotes are $\boxed{y = 2}$ and $\boxed{y = -2x}$.
Explanation
How does this work? Recall that we can think of asymptotes intuitively as "tangents at infinity". That means we may able to find asymptotes by finding the tangents at infinity. This is exactly what the above procedure does. An oblique asymptote of the form $y = mx + c$ "intersects" (or "touches") the curve $f(x, y) = 0$ "at infinity". As it is not parallel to the $y$-axis, this happens as $x \to \infty$.
To find the "intersection" of the curve $f(x, y) = 0$ with the asymptote $y = mx + c$, we proceed as usual by substituting the equation of the line in the equation of the curve. This results in an $n^{\text{th}}$ degree polynomial equation in $x$ alone (as $f(x, y)$ is a polynomial of degree $n$, in $x$ and $y$). Let this be
$a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0$.
As this is an asymptote, it must intersect the curve at only infinity, so that $\infty$ is the only root of the equation. Replace $x$ with $1/z$, and the equation becomes
$\dfrac{a_n}{z^n} + \dfrac{a_{n-1}}{z^{n-1}} + \ldots + a_0 = 0 \Rightarrow\\
a_0 z^n + a_1 z^{n - 1} + \ldots + a_{n - 1}z + a_n = 0
$
As $x \to \infty$, $z = 1/x \to 0$. But for $0$ to be a root of the equation, the constant term $a_n$ must be $0$. Then the equation becomes
$a_0 z^n + a_1 z^{n - 1} + \ldots + a_{n - 1}z = 0 \Rightarrow\\
a_0 z^{n - 1} + a_1 z^{n - 2} + \ldots + a_{n - 1} = 0
$
Again, as $0$ is the only root, we can equate $a_{n - 1}$ to $0$ as well. And so on.
Thus, $a_n = a_{n - 1} = \ldots = 0$.
Note: This is an undergraduate level explanation, and not a rigorous analytic proof.
