Theorem on the Continuity of Differentiable Functions

If a function f(x) is differentiable at a point x, then it is also continuous at that point. $$ \lim_{h \rightarrow 0} f(x)=f(x) $$ or, equivalently, $$ \lim_{h \rightarrow 0} f(x+h)=f(x) $$

However, the converse is not necessarily true.

A function being continuous at x does not automatically mean it is differentiable at x.

Proof

If a function is differentiable at x, then the limit of its difference quotient exists as h approaches 0:

$$ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $$

For the function f(x) to be continuous at x, the following condition must hold:

$$ \lim_{h \rightarrow 0} f(x+h) = f(x) $$

In other words:

$$ \lim_{h \rightarrow 0} \bigl(f(x+h) - f(x)\bigr) = 0 $$

 

Note. The formal condition for the continuity of a function is: $$ \lim_{x \rightarrow x_0} f(x) = f(x_0) $$ $$ \lim_{x+h \rightarrow x} f(x+h) = f(x) \quad \text{with} \quad x = x + h \quad \text{and} \quad x_0 = x $$ $$ \lim_{h \rightarrow x-x} f(x+h) = f(x) $$ $$ \lim_{h \rightarrow 0} f(x+h) = f(x) $$

Let’s rewrite the expression f(x+h) - f(x) in an equivalent algebraic form.

We multiply and divide by h:

$$ f(x+h)-f(x)=\frac{f(x+h)-f(x)}{h} \cdot h $$

This identity is always true: the left-hand side equals the right-hand side.

Now let’s take the limit on both sides of this equation:

$$ \lim_{h \rightarrow 0} \bigl(f(x+h) - f(x)\bigr) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \cdot h $$

By the properties of limits, we can separate the limits on the right-hand side:

$$ \lim_{h \rightarrow 0} \bigl(f(x+h) - f(x)\bigr) = \left( \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \right) \cdot \left( \lim_{h \rightarrow 0} h \right) $$

The second limit equals zero, which makes the entire right-hand side zero as well:

$$ \lim_{h \rightarrow 0} \bigl(f(x+h) - f(x)\bigr) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \cdot 0 $$

$$ \lim_{h \rightarrow 0} \bigl(f(x+h) - f(x)\bigr) = 0 $$

Since the right-hand side equals zero, it follows that the left-hand side must also equal zero.

This completes the proof that differentiable functions are continuous.

$$ \lim_{h \rightarrow 0} \bigl(f(x+h) - f(x)\bigr) = 0 $$

Why Continuity Doesn’t Necessarily Imply Differentiability

A classic counterexample is the absolute value function:

$$ f(x)=|x| $$

This function is continuous at x = 0 but is not differentiable there, because the left-hand and right-hand limits of the difference quotient do not agree.

$$ \lim_{h \rightarrow 0^-} \frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0^-} \frac{|h|}{h} = \lim_{h \rightarrow 0^-} \frac{h}{-h} = -1 $$

$$ \lim_{h \rightarrow 0^+} \frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0^+} \frac{|h|}{h} = \lim_{h \rightarrow 0^+} \frac{h}{h} = +1 $$

Therefore, in this case, the function is continuous at x = 0 but not differentiable there.

Continuity alone is not a sufficient condition for differentiability.