<s>[INST] $\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$ Help with this limit:
$$\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$$
Using L'Hopital rule gives $2m$ but I need it without using L' Hopital.
 [/INST] In the same spirit as Yves Daoust's answer, start developing  $\sqrt{1+x^2}$ using the generalized binomial theorem or Taylor series. This would give 
$$\sqrt{1+x^2}=1+\frac{x^2}{2}+O\left(x^3\right)$$ $$\sqrt{1+x^2}+x=1+x+\frac{x^2}{2}+O\left(x^3\right)$$
 $$\sqrt{1+x^2}-x=1-x+\frac{x^2}{2}+O\left(x^3\right)$$ Now,  the binomial theorem again
$$(\sqrt{1+x^2}+x)^m=1+m x+\frac{m^2 x^2}{2}+O\left(x^3\right)$$
$$(\sqrt{1+x^2}-x)^m=1-m x+\frac{m^2 x^2}{2}+O\left(x^3\right)$$ from which the result easily follows.
Just for your curiosity, if you were using the initial expansions up to $O\left(x^4\right)$, you would get 
$$(\sqrt{1+x^2}+x)^m=1+m x+\frac{m^2 x^2}{2}+\frac{1}{6} m \left(m^2-1\right) x^3+O\left(x^4\right)$$
$$(\sqrt{1+x^2}-x)^m=1-m x+\frac{m^2 x^2}{2}-\frac{1}{6} m \left(m^2-1\right) x^3+O\left(x^4\right)$$ which would make 
$$\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}=2 m+\frac{1}{3} m \left(m^2-1\right) x^2+O\left(x^3\right)$$ which will show the limit and also how it is approached.
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