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.
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.

$$\lim_{x\to-\infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\right)\frac{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}$$
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.