How do I solve $\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$?

Question

I'm having trouble finding this limit:

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

I tried multiplying by the conjugate:

$$\lim_{x\to -\infty}(\frac{\sqrt{x^2 + x + 1} + x}{1} \times \frac{\sqrt{x^2 + x + 1} - x}{\sqrt{x^2 + x + 1} - x}) = \lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x})$$

And multiplying by $\frac{\frac{1}{x}}{\frac{1}{x}}$

$$\lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}}) = \lim_{x\to -\infty}(\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1})$$

That gives me $\frac{1}{0}$. WolframAlpha, my textbook, and my estimate suggest that it should be $-\frac{1}{2}$. What am I doing wrong?

(Problem from the 2nd chapter of Early Transcendentals by James Stewart)

Answer 1

As $\;x<0\;$

$$\frac{x+1}{\sqrt{x^2+x+1}-x}\cdot\frac{-\frac1x}{-\frac1x}=\frac{-1-\frac1x}{\sqrt{1+\frac1x+\frac1{x^2}}+1}\xrightarrow[x\to-\infty]{}-\frac12$$

Answer 2

Little mistake: $$ \lim_{x\to -\infty}\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to -\infty}\frac{1 + \frac{1}{x}}{{\color{red}-}\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1}=-\frac{1}{2} $$

Answer 3

Things will be even clearer if we set $-\dfrac1x=h,$

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

$=\lim_{h\to0^+}\dfrac{\sqrt{1-h+h^2}-1}h\ \ \ \ (1)$ as $h>0$

$=\lim_{h\to0^+}\dfrac{(1-h+h^2)-1}h\cdot\dfrac1{\lim_{h\to0^+}(1-h+h^2)+1}$

$=\lim_{h\to0^+}(h-1)\cdot\dfrac1{(1-0+0^2)+1}$ cancelling $h$ as $h\ne0$ as $h\to0^+$

$=\dfrac{0-1}{1+1}$

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Alternatively, $(1)$ can also expressed $$\frac{d(\sqrt{1-x+x^2})}{dx}_{(\text{ at }x=0)}$$