Limit of square root where x approaches infinity I have to calculate the following limit, and I wondered if my solution to the question was true. Here it is:
$$\lim _{x \to -\infty} (\sqrt{(1+x+x^2)}-\sqrt{1-x+x^2})$$
Now I divide by $x^2$ and get:
$$\lim _{x \to -\infty} (\sqrt{\frac{1}{x^2}+\frac{x}{x^2}+\frac{x^2}{x^2}}-\sqrt{\frac{1}{x^2}-\frac{x}{x^2}+\frac{x^2}{x^2}})$$
I know that $$\lim_{x \to -\infty}\frac{1}{x}=0$$ so I get the following:
$$\lim _{x \to -\infty} (\sqrt{0+0+1}-\sqrt{0-0+1})$$
So we get the following:
$$\lim _{x \to -\infty} (\sqrt{1}-\sqrt{1})=0$$
Is my solution correct? 
Thanks.
 A: No, it isn't at all.
What you have shown is that
$$\lim_{x\to-\infty}\frac{\sqrt{x^2+x+1}-\sqrt{x^2-x+1}}{x}=0$$
because you have divided the function of that you were computing the limit by $x$.
To compute limits like this, it is customary to multiply and divide by an expression with the same roots, but added.
$$\lim_{x\to-\infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\right)\frac{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}$$
A: $$\sqrt{1+x+x^2}-\sqrt{1-x+x^2}=\frac{(1+x+x^2)-(1-x+x^2)}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$
$$=\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}\stackrel{(1)}=\frac{-2}{\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}+\sqrt{\frac{1}{x^2}-\frac{1}{x}+1}}$$  
$$\stackrel{x\to -\infty}\to \frac{-2}{\sqrt{0+0+1}+\sqrt{0-0+1}}=-1$$   
$(1)\!\! :\,$ I divide numerator and denominator by $|x|=\sqrt{x^2}$ and assume $x<0$, which I can do because we only care about what happens when $x\to -\infty$.
A: No. First, you divided by $x^2$ inside the square roots, which means you divided the whole expression by $|x|$. So you have shown that the difference of the square roots, divided by $x$, goes to zero.
Here's a hint: Multiply and divide the expression by the sum of the two square roots.
