How to prove this result on limit? $$\lim_{x \to 0}  \frac{({a+x^m})^{\frac{1}{n}}-({a-x^m})^{\frac{1}{n}}}{x^m}=\frac{2}{n}a^{\frac{1}{n}-1}$$
I have tried it using L'Hôpital's rule, but I don't get any way out of that.
Any Help will be appreciated. Thanks.
 A: Let $t=x^m$. Now L'Hospital's Rule will work very nicely. (It works with the original form too, but looks messier.)
Note that the derivative of $(a-t)^{1/n}$ is $-\frac{1}{n}(a-t)^{1-1/n}$. 
A: Putting $y=\frac {x^m}a,$
$$\lim_{x \to 0}  \frac{({a+x^m})^{\frac{1}{n}}-({a-x^m})^{\frac{1}{n}}}{x^m} $$
$$= a^{\frac1n}\lim_{y \to 0} \frac{(1+y)^{\frac{1}{n}}-(1-y)^{\frac{1}{n}}}{ay} $$
$$= a^{\frac1n-1}\lim_{y \to 0} \frac{1+\frac yn+O(y^2))-(1-\frac yn+O(y^2)}y \text{ (using Binomial Expansion)}$$
$$=  a^{\frac1n-1} \lim_{y \to 0}\frac {\frac {2y}n+O(y^2)}y$$
$$=\frac{2 a^{\frac1n-1}}n \text{ as }y\ne0 \text{ as }y\to0$$
A: I dont think you need to use L'Hospital's Rule at all.Its a simple consequence of mean value theorem.
By mean value theorem applied to $(a+y)^{\frac{1}{n}}$ we have
$\displaystyle\frac{({a+x^m})^{\frac{1}{n}}-({a-x^m})^{\frac{1}{n}}}{x^m}=2\frac{1}{n}(a+\epsilon)^{(1/n-1)}$, for some $\epsilon \in (-x^m,x^m)$
$\displaystyle\lim_{x \to 0} \frac{({a+x^m})^{\frac{1}{n}}-({a-x^m})^{\frac{1}{n}}}{x^m}=\lim_{x \to 0} \frac{2}{n}(a+\epsilon)^{(1/n-1)}$, for some $\epsilon \in (-x^m,x^m)$
Now by squeeze theorem we have ,
$\displaystyle\lim_{x \to 0} \frac{2}{n}(a+\epsilon)^{(1/n-1)}=\frac{2}{n}(a)^{(1/n-1)}$ as $-x^{m}<\epsilon <x^m$
