Proof that a limit is a partial derivative I'm writing up several proofs for myself, all of which have a particular sticking point.

Essentially, I want to prove that for a function $f$ of two real variables, we have

$\lim_{h \to 0} \frac{f(x \;+\; h, \; y \;+\; h) \; - \; f(x, \; y \;+ \;h)}{h} = \frac{\partial f}{\partial x}$.

When I've seen this crop up in textbooks and lecture notes (say, in the context of proving the multivariable chain rule), it's generally just been taken for granted, rather than proven.

Intuitively, how can we be sure that the function is sufficiently well behaved that having $y + h$ instead of $y$ in the second argument doesn't 'mess things up' and not allow this limit to be the partial derivative? And moreover, how could I give a rigorous proof of this?

Thanks all!
A sufficient condition for the claim to be true is that $f'_x=\partial f/\partial x$ exists in a neighbourhood of $(x,y)$ and is continuous at $(x,y)$.

Proof. Since $f'_x$ exists, the mean value theorem for derivatives says that
$$
f(x+h,y+h) - f(x,y+h) = f'_x(x+\theta h, y+h) \cdot h
$$
for some $\theta \in [0,1]$ (which may depend on $h$).
So your quotient is
$$
\frac{f(x+h,y+h) - f(x,y+h)}{h} = f'_x(x+\theta h, y+h)
,
$$
which tends to $f'_x(x,y)$ as $h \to 0$, because of the assumption that $f'_x$ is continuous there.
There's no way anything could be messed up since the second coordinate is always fixed. It's only the first that's varying. That is, we're considering the change from the point $(x,y+h)$ to the point $(x+h,y+h).$ Throughout this change, the second coordinate is clearly fixed. Thus, it's not concerned with the limiting process at all. This is why we can be sure that the limit, if it exists, is indeed the derivative with respect to $x,$ fixing $y.$ Thus, there's nothing more to justify than in the usual derivative. Hence, what you noticed about the books.