Proof that a function in $\mathbb{R}^2$ is differentiable at a point $a$. I struggle with a proof in multivariable calculus:
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a continuous function that is differentiable at $\mathbb{R}^2\backslash \{a\}$ with $a\in\mathbb{R}^2$. The limit $\lambda:=\lim_{x\rightarrow a}\nabla f(x)$ exists. Show that $f$ is differentiable at the point $a$, also show that $\nabla f(a)=\lambda$.
I think this statement can be proven by using the Mean Value Theorem for multiple variables, but I don't know how to do it. I would appreciate your help.
 A: Note that
\begin{eqnarray}
f(a+h)-f(a) - \lambda h &=& f((a_1+h_1,a_2+h_2))-f((a_1,a_2+h_2))  \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  +f((a_1,a_2+h_2)) - f(a) - \lambda h \\
&=& {\partial f(a_1+\xi_1,a_2+h_2) \over \partial x_1}h_1 + {\partial f(a_1,a_2+\xi_2) \over \partial x_2}h_2 - \lambda h
\end{eqnarray}
for some $\xi_1 \in (a_1,a_1+h_1), \xi_2 \in (a_2,a_2+h_2)$.
Since ${\partial f(x) \over \partial x} \to \lambda$, for any $\epsilon>0$ we can find some $\delta>0$ such that
$\|{\partial f(a+h) \over \partial x} - \lambda\| <\epsilon$ for all
$\|h\| < \delta$. Then the above shows that
$\|f(a+h)-f(a) - \lambda h \| \le \epsilon \|h\|$. Hence $f$
is differentiable at $a$ with derivative $\lambda$.
Alternative:
Note that ${\partial f(x) \over \partial x}$ is bounded in a punctured neighbourhood of $a$, and so is bounded on any segment of the form $(a,x)$ (that is, the line segment not including end points) as long as $x$ is in this punctured neighbourhood.
Hence for any sufficiently small $h$, the function $\phi(t)=f(a+th)$ is absolutely continuous and so
$\phi(1) = \phi(0) + \int_0^1 \phi'(t) dt$.
Hence $f(a+h) - f(a) - \lambda h = \int_0^1 ({\partial f(a+th) \over \partial x}-\lambda)dt h$.
Let $\epsilon >0$ and choose $\delta>0$ such that
$\|{\partial f(x) \over \partial x}-\lambda\| < \epsilon$ whenever 
$\|h\| < \delta$. Then
$\|f(a+h)-f(a) - \lambda h \| \le \epsilon \|h\|$.
