$\lim_{x \to 0}\frac{(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m}{x}$ 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.
 A: hint: Use the formula: $a^m - b^m = (a-b)(a^{m-1}+ a^{m-2}b + a^{m-3}b^2+\cdots + b^{m-1})$, and the limit just comes out nicely.
A: Hint. One may recall that, for any differentiable function $f$ near $0$, one has
$$
\lim_{x \to 0}\frac{f(x)-f(0)}{x}=f'(0)
$$ then apply it to
$$
f(x)=(\sqrt{1+x^2}+x)^m-(\sqrt{1+x^2}-x)^m
$$ giving$$
f'(0)=2m.
$$
A: Using the identity:
$$a^m - b^m = (a - b)(a^{m - 1} + a^{m - 2}b + \cdots + b^{m - 1}).$$
The operand equals to 
$$2\left[(\sqrt{1 + x^2} + x)^{m - 1} + \cdots + 
(\sqrt{1 + x^2} - x)^{m - 1}\right] \to 2(1 + \cdots + 1) = 2m$$
as $x \to 0$.
A: If you develop the numerator using the Binomial formula, you will get two terms $(\sqrt{1+x^2})^m$ which cancel each other, then two terms $mx(\sqrt{1+x^2})^{m-1}$ which add, and other terms with higher powers of $x$.
After simplification by $x$, the limit is $2m$.
A: 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.
