# Double limit in the proof of differentiability

A sufficient condition for differentiability of a function $f$ about a point $(x_0,y_0)$ is that the partial derivatives exist in a neighborhood of the point and they are continuous at the point. To prove this, Thomas calculus uses the linearization definition of differentiable function namely $f(x_0+h,y_0+k)=f(x_0,y_0)+hf_x(x_0,y_0)+kf_y(x_0,y_0)+h\epsilon_1+k\epsilon_2$ where $\epsilon_1\to0$ and $\epsilon_2\to0$ as $(h,k)\to(0,0)$. (See appendix 7 in 11th edition for the proof)
My question is on one step used in this proof.

The proof they use is given below:
$f(x_0+h,y_0+k)-f(x_0,y_0)=f(x_0+h,y_0+k)-f(x_0+h,y_0)\ \ +f(x_0+h,y_0)-f(x_0,y_0)$. Now the function $G$ defined on the interval with end points $y_0,y_0+k$ as $G(y)=f(x_0+h,y)$ is differentiable and hence continuous with derivative $G'(y)=f_y(x_0+h,y)$. By applying the mean value theorem on $G$, we obtain a value $d$ between $y_0$ and $y_0+k$ at which $G(y_0+k)-G(y_0)=k G'(d)$ or $f(x_0+h,y_0+k)-f(x_0+h,y_0)=k f_y(x_0+h,d)$.

Similarly the function $F(x)=f(x,y_0)$ is differentiable and hence continuous in an interval with end points $x_0,x_0+h$ and $F'(x)=f_x(x,y_0)$. Thus by appling MVT on $F$, we obtain a point $c$ between $x_0$ and $x_0+h$ at which $F(x_0+h)-F(x_0)=h F'(c)$ or $f(x_0+h,y_0)-f(x_0,y_0)=h f_x(c,y_0)$

The next is the part that is not clear to me.

Now, as both $h$ and $k\to0$, we know that $c\to x_0$ and $d\to y_0$. Since $f_x$ and $f_y$ are continuous at $(x_0,y_0)$, the quantities $\epsilon_1=f_x(c,y_0)-f_x(x_0,y_0)$ and $\epsilon_2=f_y(x_0+h,d)-f_y(x_0,y_0)$ both approach zero as $h$ and $k\to0$.

Once we get $\epsilon_1\to0$ and $\epsilon_2\to0$ as $(h,k)\to(0,0)$, we are through. It is obvious that $\epsilon_1$ doesn't depend $k$ so that $\epsilon_1\to0$ as $(h,k)\to(0,0)$. But $\epsilon_2$ depends on both $h$ and $k$. What we have directly is the iterated limit $\lim_{h\to0}\lim_{k\to0}\epsilon_2$. I am not quite convinced that the double limit $\boldsymbol{\lim_{(h,k)\to(0,0)}\epsilon_2}$ exists. Don't we need to show that it exists and the limit is indeed 0? Isn't the proof in the text a little sloppy there? Any kind of help is appreciated. Thank you.

## 1 Answer

The limit exists because $f_y$ is continuous. In particular, the fact that $f_y$ is continuous at $(x_0,y_0)$ means that

$$\lim_{(x,y)\to(x_0y,_0)} f_y(x,y) = f_y(x_0,y_0).$$

To see this, suppose for the moment that $d \to y_0$ as $(h,k) \to (0,0)$. The continuity of $f_y$ then guarantees that

$$\varepsilon_2 = f_y(x_0+h,d)-f_y(x_0,y_0) \to 0 \quad \text{as} \quad (h,k)\to(0,0).$$

Hence, all that we need to establish is that $d$ does in fact approach $y_0$. Recall that $d$ lies in $(y_0,y_0+k)$ and satisfies

$$f(x_0+h,y_0+k)-f(x_0+h,y_0) = kf_y (x_0+h,d).$$

You are correct to point out that we need to be a bit careful here, since $d$ depends on both $h$ and $k$. However, the fact that $d \in (y_0,y_0+k)$ for any $k$ means that for every neighbourhood $U$ of $y_0$, I can choose a corresponding neighbourhood $V$ of $(0,0)$ such that whenever $(h,k) \in V$, we must have that $d \in U$. That is to say, $d \to y_0$ as $(h,k) \to (0,0)$.

(I omitted the details of implicit $\varepsilon$-$\delta$ arguments in the discussion above since I don't think they add that much more clarity to the answer. However, I hope I convinced you that they can be done, and possibly helped you see how you might execute them if you were so inclined.)

• But we first fixed $h$ and using it defined a function $G$ and then applied MVT to $G$ to obtain a point $d$. Now how can we let $(h,k)$ approach (0,0) in any direction? – Cyriac Antony Apr 11 '18 at 10:58
• @CyriacAntony All we require is that $\varepsilon_2 \to 0$ as $(h,k) \to (0,0)$. In order for this to hold, it is sufficient for $x_0 + h \to x_0$ and $d \to y_0$ as $(h,k)\to (0,0)$. $d$ is a function of $h$, but since $d$ must lie in $(y_0,y_0+k)$ for all $(h,k)$, we must have that $d \to y_0$ as $(h,k)\to (0,0)$. – Theoretical Economist Apr 11 '18 at 11:35