$\limsup_{x\to 0}$ and $\liminf_{x\to 0}$

Question

I want to find $\lim_{x→0}\sqrt{1+x+x²}=1$ and want to show that $\sqrt{1+x+x²}-1/(\sqrt{1+x}-\sqrt{1-x})$ tends to a limit as $x\to 0$

So in the first case I want to show that the $\limsup_{x\to 0}$ and $\liminf_{x\to 0}$ are the same but I don't understand what $\limsup_{x\to 0}$ and $\liminf_{x\to 0}$ are. So I'm guessing that it might be smarter to substitute $x$ with $1/y$ and let $y\to \infty$, but am I allowed to do this?

for the second part I guess I'd just multiply by $\sqrt{1+x}+\sqrt{1-x}$ and use the first result?

Answer 1

Note that continuous functions commute with limits: https://proofwiki.org/wiki/Limit_of_Image_of_Sequence.

Therefore we have

$$\lim_{x\to 0}\sqrt{1+x+x²} = \sqrt{\lim_{x\to 0}\left( 1 + x +x^2\right)}=\sqrt{1}=1.$$

Your reasoning for the second part seems correct.