# What is the difference between continuity and differentiability of functions of 2 or more variables, intuitively?

I was reading my calculus book. For continuity, I was first told to imagine a surface to be a mountain; where there is a large rock, for ex, whose side rise vertically upward, or where there is a sudden change in altitude, there is dis-continuity. Another case of dis-continuity is where there is a hole, or chasm, like the one in the graph of $z=\frac{-1}{(x-y)^2}$. I was confused. I thought those places of sudden changes in altitude are places of non-differentiability (I was taught that for functions of 1 variable). The book then says in symbols, continuity is (say there is 2 points $P(x_0, y_0, z_0)$ and $Q(x, y, z)$), $lim_{x\rightarrow x_0, y\rightarrow y_0}f(x,y)=f(x_0,y_0)$. But I thought this limit expression doesn't say anything about the things like sudden changes in alitutide.

Then I got to part about differentiability. Here, the book explains that a surface may indeed be continuous but may contain crags, which have sharp points or edges. This is where I suddenly realize the striking, yet confusing resemblance between what continuity is and what differentiability is. Does sharp points just places of sudden change in altitude (z) or other variables?

If a function $f: A \subset \Bbb R^n \to \Bbb R$, where $A\neq \varnothing$ is an open set, is discontinuous at a point, it's automatically non-differentiable at the point. Continuity is a necessary condition for differentiability; one first restricts their attention to continuity, then worries about differentiability.
For $n=1$, such a function, intuitively, is continuous at a point if it gets arbitrarily close to its value at this point when $x$ gets near the point from either directions; in other words, it doesn't have a "jump" at the point neither from the right nor from the left. For $n=2$, the concept is nearly the same, however we are concerned here with infinitely many directions through which the point might be approached, so for example, a function might be continuous at $(0,0)$ if one restricts the attention to the paths $y = 0$ and $x=0$, however discontinuous otherwise. Consider, for instance, $f(x,y) = \frac{xy}{x^2 + y^2}$ if $(x,y) \neq (0,0)$ and $0$ otherwise. If one computes $\lim_{(x,y) \to (0,0)} f(x,y)$ over the paths $x=0$ and $y=0$, one gets $0$. However, over the path $y = x$, the limit is $1/2$. To sum up, for $n=2$, a function is continuous at a point if it gets arbitrarily close to its value at the point as $(x,y)$ approaches the point from any possible direction.
Roughly speaking, points of non-differentiability of a continuous function, for $n=1$, are points where the graph has a "corner", and a "corner" is a point where the function "suddenly" changes its "trajectory", but still doesn't "jump" (imagine the trajectory of a diagonally thrown ball and the point at which it hits the ground vs the trajectory of a skateboarder over a skate-park). For $n=2$, it is rather a set of points over which the surface doesn't "flow smoothly" or has an "edge" (imagine a "smooth flow" as that of the surface of a skate-park or a saddle). This is because, intuitively, the derivative of $f$ should give information about the "rate of change" of the function. So for instance, consider two planes which intersect diagonally at a line $\ell$. The upper part corresponds to a function $f$ which is continuous but non-differentiable over the set of points $(x,y)$ for which $f(x,y) \in \ell$ (a concrete example would be $f(x,y) = x$ for $x\ge 0$ and $-x$ for $x < 0$ - the function is continuous on $\Bbb R^2$ and non-differentiable over the $y$-axis).