Why differentiability implies continuity, but continuity does not imply differentiability?

Solution 1:

Here is an intuitive explanation.

Continuity requires that $f(x)-f(y)\to 0$ as $x - y \to 0$.

Differentiability requires that $f(x)-f(y)\to 0$ as $x - y \to 0$, and that $f(x)-f(y)\to 0$ at least as fast as $x - y \to 0$ (in the sense that the ratio still has a limit).

In particular, the conditions for differentiability include the condition for continuity.

Solution 2:

Why does differentiability imply continuity, but continuity does not imply differentiability?

Why are all differentiable functions continuous, but not all continuous functions differentiable? For the same reason that all cats are animals, but not all animals are cats. Differentiability also implies a certain “smoothness”, apart from mere continuity. It is perfectly possible for a line to be unbroken without also being smooth. Fractals, for instance, are quite “rugged” $($see first sentence of the third paragraph: “As mathematical equations, fractals are usually nowhere differentiable”$)$.

Solution 3:

Differentiability guarantees continuity in the following way :

Consider a function $f:R \rightarrow R$ which is differentiable at a point $x_0 \in R$ and it’s derivative is $L$. Thus by first principle derivative we have the RHD as

$$\lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h} = L$$ $$\implies \lim_{h\to0} [f(x_0 + h) - f(x_0)] = \lim_{h\to0} hL$$ $$\implies \lim_{h\to0} [f(x_0 + h) - f(x_0)]= 0$$

Which is the condition of continuity on the right hand side

We can prove the same for left hand side as well

Therefore differentiability always guarantees continuity but not the vice versa since the LHD May not be = RHD when trying to reverse engineer the above