Solving $\lim_{x\to -\infty}{\sqrt{x^2+x+1} - \sqrt{1-x+x^2}}$ using $o$ ( little - oh )

I was solving the following limit: $$\lim_{x\to -\infty}{\sqrt{x^2+x+1} - \sqrt{1-x+x^2}}$$

The limit itself is not too difficult ( expand to a fraction and just solve it, remembering the $-$ ) but the solution given in the manual is what troubles me. The solution goes as follows: $${\sqrt{x^2+x+1} - \sqrt{1-x+x^2}} = -x\bigg({\sqrt{1+\frac{1}{x}+\frac{1}{x^2}} - \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}}\bigg)$$ $$-x\bigg({\sqrt{1+\frac{1}{x}+\frac{1}{x^2}} - \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}}\bigg) = -x\bigg[ \bigg( 1 + \frac{1}{2}\big(\frac{1}{x}+\frac{1}{x^2}\big) + o\big( \frac{1}{2}\big(\frac{1}{x}+\frac{1}{x^2}\big) \big) - \bigg( 1 + \frac{1}{2}\big(-\frac{1}{x}+\frac{1}{x^2}\big) + o\big( \frac{1}{2}\big(-\frac{1}{x}+\frac{1}{x^2}\big) \big) \bigg) \bigg]$$

Which of course is $-1$ as $x$ approaches $-\infty$. While i understand the $o$ notation , I still have a questions here:

• How is $${\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}} = 1 + \frac{1}{2}\big(\frac{1}{x}+\frac{1}{x^2}\big) + o\big( \frac{1}{2}\big(\frac{1}{x}+\frac{1}{x^2}\big) \big)$$ when $x\to -\infty$.

HINT:

Let $-1/x=h$ $$x^2+x+1=\dfrac{1+h+h^2}{h^2},\sqrt{1+x+x^2}=\dfrac{\sqrt{1+h+h^2}}{\sqrt{h^2}}$$

Now as $h>0,\sqrt{h^2}=|h|=+h$

$$\lim_{x\to -\infty}{\sqrt{x^2+x+1} - \sqrt{1-x+x^2}} =\lim_{h\to0^+}\dfrac{\sqrt{1+h(h-1)}-\sqrt{1+h(h+1)}}h$$

Now using Taylor's Expansion, $$\sqrt{1+h(h-1)}=\{1+h(h-1)\}^{1/2}=1+\dfrac{h(h-1)}2+o(h^2)$$

$$\sqrt{1+h(h+1)}=\{1+h(h+1)\}^{1/2}=1+\dfrac{h(h+1)}2+o(h^2)$$

If Taylor's Expansion is not mandatory, $$\sqrt{1+h(h-1)}-\sqrt{1+h(h+1)}=\dfrac{1+h(h-1)-\{1+h(h+1)\}}{\sqrt{1+h(h-1)}+\sqrt{1+h(h+1)}}=\dfrac{-2h}{\sqrt{1+h(h-1)}+\sqrt{1+h(h+1)}}$$

• Aha! That's it. The problem with this is that these problems come up before the Taylor's expansion is covered in the book, so i thought there's something else behind this. The second method is the one I did. Thanks. – Transcendental Dec 13 '16 at 12:44
• – lab bhattacharjee Dec 13 '16 at 12:53