I have to calculate the following limit
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x$$
it is in un undeterminated form.
I tried to rewrite it as follows:
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - \sqrt{|x|^2}$$
but seems a dead road.
Can anyone suggest a solution?
thanks for your help
Try rationalizing the expression. That is, multiply by $1=\frac{\sqrt{x^2+2x+2}+x}{\sqrt{x^2+2x+2}+x}$ to get $$ \frac{(x^2+2x+2)-x^2}{\sqrt{x^2+2x+2}+x} $$ What can we do from there?
HINT:
Putting $y=-x,$ $$ \sqrt{x^2+2x+2} - x=\sqrt{y^2-2y+2} +y= \frac{(\sqrt{y^2-2y+2} +y)(\sqrt{y^2-2y+2} -y)}{(\sqrt{y^2-2y+2} -y)}$$ $$=\frac{y^2-2y+2-y^2}{\sqrt{y^2-2y+2} -y}=\frac{-2+\frac2y}{\sqrt{1-\frac2y+\frac2{y^2}}-1}$$
Clearly $$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x=+\infty+\infty=+\infty$$ But \begin{gather*}\lim_{x\rightarrow +\infty} \sqrt{x^2+2x+2} - x="\infty-\infty"=\\ =\lim_{x\rightarrow +\infty} \frac{(\sqrt{x^2+2x+2} - x)(\sqrt{x^2+2x+2} + x)}{\sqrt{x^2+2x+2} + x}=\lim_{x\rightarrow +\infty} \frac{2x+2}{\sqrt{x^2+2x+2} + x}=\lim_{x\rightarrow +\infty} \frac{2+2/x}{\sqrt{1+2/x+2/x^2} + 1}=1 \end{gather*}
Assuming you meant $\sqrt{x^2+2x+2} + x$ (as $\sqrt{x^2+2x+2} - x \to +\infty$ when $x\to-\infty$):
Another option would be to use asymptotics and known Taylor expansions (at $0$): for $x\to-\infty$, $$ \begin{align*} \sqrt{x^2+2x+2} + x &= |x|\sqrt{1+\frac{2}{x}+\frac{2}{x^2}} - |x| \\ &= |x|\left( 1+\frac{1}{2}\cdot\frac{2}{x} + o\left(\frac{1}{x}\right) - 1 \right) \\ &= \frac{|x|}{x} + o(1) \sim -1 \end{align*} $$ so the limit is $-1$.
