How to evaluate the limit of $\sqrt{1+x+x^2}?$

Question

I’ve got a short question. Why does $$\lim_{x\to 0} \sqrt{1+x+x^2} \text{ behave as } \lim_{x\to 1} \sqrt x \ ?$$

Answer 1

Note that by continuity

$$\lim_{x\to 0} \sqrt{1+x+x^2}=\sqrt{1+0+0^2}=1$$

$$\lim_{x\to 1} \sqrt x=\sqrt 1=1$$

and since

$$\lim_{x\to 1} \sqrt x=\lim_{x\to 0} \sqrt {1+x}=1$$

then

$$\lim_{x\to 0} \frac{\sqrt{1+x+x^2}}{ \sqrt {1+x}}=\sqrt{1+\frac{x^2}{1+x}}=1$$

From the latter we say that for $x\to0$

$$\sqrt{1+x+x^2}\sim \sqrt {1+x}$$

then, by definition, $\sqrt{1+x+x^2}$ and $\sqrt {1+x}$ are asymptotic equivalent at the neighborhood of $x=0$ and since the behaviour of $\sqrt {1+x}$ at $x=0$ is equivalent to the behaviour of $\sqrt {x}$ at $x=1$, in this sense, we can state that $\lim_{x\to 0} \sqrt{1+x+x^2}$ behaves as $\lim_{x\to 1} \sqrt x$.