<s>[INST] Finding a limit results in division by 0. $$
\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x^2}.
$$
 Tried multiplying by $\sqrt{1+x}+1$ but got $1/(x(\sqrt{1+x}+1))$ where substitution results in $1/0$ which is illegal (or maybe $1/0$ is Infinity?). Second approach is to divide by $x^2$ which leads to the $\sqrt{1/x^2 +1)}- 1/x^2$. How to solve it?
 [/INST] $$\lim_{x\to 0 }\frac{\sqrt{x+1}-1}{x^2}=\lim_{x\to 0 }\frac{\sqrt{x+1}-1}{x^2}\cdot 1=\lim_{x\to 0 }\frac{\sqrt{x+1}-1}{x^2}\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$$
$$=\lim_{x\to 0 }\frac{x+1-1}{x^2(\sqrt{x+1}+1)}=\lim_{x\to 0 }\frac{x}{x^2(\sqrt{x+1}+1)}$$
$$=\lim_{x\to 0 }\frac{1}{x(\sqrt{x+1}+1)}$$
If $x\rightarrow 0^- \Rightarrow \frac{1}{x(\sqrt{x+1}+1)}\rightarrow -\infty$, and
$x\rightarrow 0^+ \Rightarrow \frac{1}{x(\sqrt{x+1}+1)}\rightarrow +\infty$
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