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

For some reason I fail to evaluate this (apparently) simple limit:

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

I tried conjugate multiplication* however it didn't work for me. I thought about sandwiching, but I don't see how to do it here. I was also trying to evaluate it as a composition of functions but with no luck. Any suggestions?

*

Conjugate multiplication gives

$$\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$

I tried working with this by factoring and canceling things out but it didn't work.

• Could we see your work for conjugate multiplication? What didn't work? – user223391 May 7 '15 at 19:02
• Note that, as $x \rightarrow -\infty$, you can write $\sqrt{1 + x + x^2} = -x\sqrt{1/x^2 + 1/x + 1}$ – Alex Zorn May 7 '15 at 19:06

Conjugate multiplication will help and what you got is correct.

$$\lim_{x\to -\infty}\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$$$$=\lim_{x\to -\infty}(\sqrt{1+x+x^2}-\sqrt{1-x+x^2})\cdot\frac{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$ $$=\lim_{x\to -\infty}\frac{2x}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$$ (here setting $t=-x$ gives you) $$=\lim_{t\to \color{red}{+}\infty}\frac{-2t}{\sqrt{1-t+t^2}+\sqrt{1+t+t^2}}$$ $$=\lim_{t\to\infty}\frac{-2}{\sqrt{\frac{1}{t^2}-\frac{1}{t}+1}+\sqrt{\frac{1}{t^2}+\frac 1t+1}}$$ $$=\frac{-2}{1+1}$$

• I didn't think of dividing the numerator and the denominator by $\sqrt{t^2}$. Thank you. Simple indeed. – user238281 May 7 '15 at 19:11

HINT: multiply numerator and denominator by $$\sqrt{1+x+x^2}+\sqrt{1-x+x^2}$$

• The OPer already said s/he had tried this. – Simon S May 7 '15 at 19:04

If you factor out one power of $x$ and substitute $y = 1/x$ you get $$x(\sqrt{y^2+y+1} - \sqrt{y^2-y+1})$$ It is much easier (for me) to take the limit of this expression as $y \rightarrow 0$. If you then re-substitute for $x$ you get the right answer. This is due to the properties of limits of the product of expressions: You can do an inner limit first when it makes sense.

• No. Expanding $\sqrt{u + 1}$ for small $u$ to third order gives: $1+(u/2)-(u^2/8)+(u^3/16)$ Do this for $u = y^2+y$ and for $u=y^2-y$ separately. – amcalde May 7 '15 at 20:01