Computing this limit: $ \lim_{y\to0} \frac{f(x,y) - f(x-y,y)}{y} = g(x)$

Question

If $f(x,y) \in \mathbb{R^2}$ and $g(x) \in \mathbb{R}$. Assuming $\frac{f(x,y) - f(x-y,y)}{y} = g(x); \forall y \in \mathbb{R}$ $$$$ Can we do the following: $$ \lim_{y\to0} \frac{f(x,y) - f(x-y,y)}{y} = g(x)$$ $$ \lim_{y\to0} \frac{\partial{f(x,y)}}{{\partial x}}= g(x) $$

Answer 1

No, that's not quite valid, though it's getting you to the right answer. The difference quotient inside the limit is not equal to the partial $\frac{\partial f}{\partial x}$, because it's just a difference quotient; it doesn't become a derivative until you take a limit. And even after you do, that quotient is not simply a partial in the $x$ direction, because both terms are changing.

$\begin{align*} \lim_{y \to 0} \frac{f(x,y) - f(x-y,y)}{y} &= \lim_{y \to 0} \frac{f(x,y) - f(x,0)}{y} - \lim_{y \to 0} \frac{f(x-y,y) - f(x,0)}{y} \\ &= \frac{\partial f}{\partial y}(x,0) - \left(-\frac{\partial f}{\partial x}(x,0) + \frac{\partial f}{\partial y}(x,0)\right) \\ &= \frac{\partial f}{\partial x}(x,0) \end{align*}$

For the second equality above I recognized those difference quotients as directional derivatives.

Answer 2

Why not to use Taylor expansion around $y=0$ ? $$f(x,y)=f(x,0)+y f^{(0,1)}(x,0)+O\left(y^2\right)$$ $$f(x-y,y)=f(x,0)+y \left(f^{(0,1)}(x,0)-f^{(1,0)}(x,0)\right)+O\left(y^2\right)$$ $$f(x,y)-f(x-y,y)=y f^{(1,0)}(x,0)+O\left(y^2\right)$$ $$\frac{f(x,y)-f(x-y,y) }y=f^{(1,0)}(x,0)+O\left(y^1\right)$$