Prove a bivariate function has a continuous extension Define function $g(t,p) = (1+ p (e^t -1))^{1/t}$, where $p\in [0,1], t\in [0,1/2]$. It is clear that $\lim_{t\to 0} g(t,p) = e^p = g(0,p)$. 
Define $h(t,p) = \frac{g(t,p) - g(0,p)}{t}$. Prove the following statement: 
The bivariate function
\begin{align}
\frac{\partial h(t,p)}{\partial t}
\end{align}
can be extended as a continuous function on $(t,p)\in [0,1/2]\times [0,1]$. 
 A: For $t>0$, the formula for $h(t,p)$ is well defined and we have that
$$h(t,p)=\frac{{\left(1+p\left(e^t-1\right)\right)}^\frac{1}{t}-e^p}{t}$$
In that range, $h(t,p)$ is differentiable and we have that $\frac{\partial}{\partial t}h(t,p)$ equals
$$\frac {\Big[- \left( p{{\rm e}^{t}}-p+1 \right) ^{{t}^{-1}}\ln  \left( p{
{\rm e}^{t}}-p+1 \right) {{\rm e}^{t}}p+{{\rm e}^{p+t}}pt+ \left( p{
{\rm e}^{t}}-p+1 \right) ^{{t}^{-1}}\ln  \left( p{{\rm e}^{t}}-p+1
 \right) p + \left( p{{\rm e}^{t}}-p+1 \right) ^{{t}^{-1}}pt-{{\rm e}^{
p}}pt- \left( p{{\rm e}^{t}}-p+1 \right) ^{{t}^{-1}}\ln  \left( p{
{\rm e}^{t}}-p+1 \right) - \left( p{{\rm e}^{t}}-p+1 \right) ^{{t}^{-1
}}t+{e}^pt\Big]}{t^3 \left( p{e}^t-p+1 \right) }
$$
It suffices to show that $\frac{\partial}{\partial t}h(t,p)$ is bounded near $t=0$. This can be done by calculating the limit $\lim_{t\to0}\frac{\partial}{\partial t}h(t,p)$ in the previous expression, through repeated applications of L'Hôpital, although the specific caluclations become rather involved.
I used Maple and found the limit to be
$$\lim_{t\to0}\frac{\partial}{\partial t}h(t,p)=\frac{1}{24}\,{{\rm e}^{p}}p \left( 3\,{p}^{3}+2\,{p}^{2}-9\,p+4 \right) $$
