Let $f(x,y)$ be a real-valued function in two variables.
Let $x_n \to x_0$ and $y_n \to y_0$ as $n \to \infty$.
We assume $f(x,y)$ is differentiable in $x$ at $(x_0, y_0)$ with derivative $f'_x(x_0,y_0)$. Further suppose $f'_x(x,y)$ is continuous at $(x_0,y_0)$.
Question: Consider the following Taylor expansion about $(x_0,y_n)$:
$$f(x_n,y_n) = f(x_0,y_n) + f'_x(x_0,y_n)(x_n – x_0) + R_n(x_n,y_n)$$Is it neccesary true that $R_n(x_n,y_n) = o(|x_n – x_0|)$?
Clearly if all instances of $y_n$ are replaced by $y_0$, the Taylor expansion is correct based on the definition of differentiability. And if $f(x,y)$ is differentiable in both $x$ and $y$ at $(x_0,y_0)$, we have
$$f(x_n,y_n) = f(x_0,y_0) + f'_x(x_0,y_0)(x_n – x_0) + f'_y(x_0,y_0)(y_n – y_0) + R_n^*(x_n,y_n)$$
with $R^*_n(x_n,y_n) = o(|x_n – x_0| + |y_n – y_0|)$.
However, if we don't have differentiability in the two variables, I wonder if the Taylor expansion in the question is correct.
Best Answer
Counterexample: Define $f$ to be $0$ on the set $\{y>|x|\}\cup \{(x,y):y\le 0\},$ and $f=1$ everywhere else. Let $(x_0,y_0)=(0,0).$ Then $f_x(0,y)=0$ for all $y.$ But $f(1/n,1/(2n)) = 1$ for all $n,$ which forces $R_n(1/n,1/(2n)) = 1$ for all $n.$