I have this problem that I can't solve :
$$\lim_{x\to 1} \frac{\dfrac{1}{(8(x^{1/2^n}+h^4))^{0.25}}+\dfrac{x^{1/2^n}}{h+1} + \dfrac{h}{x^{1/2^n}+1}- \left(\dfrac{1}{(8(x^{1/2^{n+1}}+h^4))^{0.25}} + \dfrac{x^{1/2^{n+1}}}{h+1}+\dfrac{h}{x^{1/2^{n+1}}+1} \right)}{x^{1/2^n}-x^{1/2^{n+1}}}$$
If we treat element by element it's easy but I would like to have another solution.
You can use the L'hospithal rule.
Given $$ g(x,n) = {{\Delta _n \,f(x,n)} \over {\Delta _n \,x^{\,1/2^{\,n} } }} \quad \quad f(x,n) = {1 \over {\left( {8\left( {x^{\,1/2^{\,n} } + h^{\,4} } \right)} \right)^{\,1/4} }} + {{x^{\,1/2^{\,n} } } \over {h + 1}} + {h \over {x^{\,1/2^{\,n} } + 1}} $$ you are, possibly, looking for something like this $$ \eqalign{ & \mathop {\lim }\limits_{x \to 1} g(x,n) = {{\Delta _n \,\mathop {\lim }\limits_{x \to 1} f(x,n)} \over {\Delta _n \,\mathop {\lim }\limits_{x \to 1} x^{\,1/2^{\,n} } }} = {{\Delta _n \,\mathop {\lim }\limits_{dx \to 0} f(1 + dx,n)} \over {\Delta _n \,\mathop {\lim }\limits_{dx \to 0} \left( {1 + dx} \right)^{\,1/2^{\,n} } }} = \cr & = \mathop {\lim }\limits_{dx \to 0} {{\Delta _n \,\left( {f(1,n) + f'(1,n)dx} \right)} \over {\Delta _n \,\left( {1 + {1 \over {2^n }}dx} \right)}} = \mathop {\lim }\limits_{dx \to 0} {{\,\Delta _n f(1,n) + \Delta _n f'(1,n)dx} \over { - {1 \over {2^{n + 1} }}dx}} = \cr & = - 2^{n + 1} \Delta _n f'(1,n) \cr} $$ since $$ \Delta _n f(1,n) = 0 $$
It turns out that $$ \eqalign{ & \mathop {\lim }\limits_{x \to 1} g(x,n) = - 2^{n + 1} \Delta _n f'(1,n) = \cr & = {1 \over {h + 1}} - {1 \over {2^{\,11/4} \left( {h^{\,4} + 1} \right)^{\,5/4} }} - {h \over 4} \cr} $$ and it does not depend on $n$.