Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
 A: $1 - x + \sqrt{2 + 2x + x^2} = \dfrac{(1-x)^2 - (2 + 2x + x^2)}{1 - x - \sqrt{2 +2x + x^2}} = \dfrac{-1 - 4x}{1 - x - \sqrt{2 + 2x + x^2}} = \dfrac{-\dfrac{1}{x} - 4}{\dfrac{1}{x} - 1 - \sqrt{\dfrac{2}{x^2} + \dfrac{2}{x} + 1}} \to \dfrac{-4}{-1-1} = 2$ when $x \to \infty$
A: You should note that you get
$$\sqrt{x^2 + 2x + 2} = \sqrt{(x+1)^2 + 1} \approx (x+1)$$
as you are squaring i.e. $(x+1)^2$ is not approximately equal to $x^2$  since $(x+1)^2 - x^2  \approx 2x$ you can only ignore the constant.
A: You need to Taylor expand the square root part as well:
Recall that $\sqrt{1+x} = 1 + \frac{1}{2}x + o(x)$ for $x \to 0$
$\sqrt{2/x^2+2/x+1} = (1+\frac{1}{2}2/x+o(1/x)) = 1+1/x + o(1/x)$ as $x \to \infty$
Now you get:
$1-x+x\sqrt{2/x^2+2/x+1} = 1-x+x[1+1/x + o(1/x)] = 2 + o(1) \to 2$ as $x \to \infty$
A: You have
\begin{align}
\lim_{x\rightarrow\infty}\left(1-x+\sqrt{2+2x+x^2}\right)
    &=\lim_{x\rightarrow\infty}\left(1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}\right) \\
    &= \lim_{x\to\infty} \left(1+x\left(\sqrt{\frac{2}{x^2}+\frac{2}{x}+1}-1\right)\right).
\end{align}
The second term is of the form $\infty\cdot 0$; you cannot conclude that it goes to zero (and, in fact, it goes to $1$) as $x\to\infty$.
A: Note that 
$$\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}\rightarrow 0$$ as $x\rightarrow \infty$
Your expression is equal to 
$$\sqrt{(x+1)^2+1}-(x+1)+2 \rightarrow 2$$ as $x\rightarrow \infty$.
A: For convention, let $\lim$ denote $\lim_{x\rightarrow \infty}$. I avoid big-O notation and power series here and perhaps it is clearer this way?
Now, $$\lim [1-x+\sqrt{2+2x+x^2}] \\
= \lim [1 + \frac{2+2x}{x + \sqrt{2+2x+x^2}}] \\
= \lim [1 + \frac{2+2x}{x(1+\sqrt{2/x^2+2/x+1})}] \\
= \lim [1 + \frac{2/x + 2}{1+\sqrt{2/x^2+2/x+1}}] \\
= 1 + \frac{2}{1+\sqrt{1}} = 2.$$
