We will now deal with the connection between continuity and differentiability of a function. So, what can we say about the differentiability of a continuous function? And furthermore, what can we say about the continuity of a differentiable function? To answer these questions, we will only focus on continuous or differentiable functions at a given point \( x_0 \) on the function’s domain.
The key to understand these concepts is the notion of the derivative of a function seen as the limit of the difference quotient as \( \Delta x \) tends to zero. Furthermore, we will make use of the definition of continuity as seen on limits theory in order to understand the connection between differentiability and continuity.
Le’ts now see the connection between differentiability and continuity of a function at a given point \( x_0 \) on the domain, linking together the notions of differentiability and continuity.
Continuity of a function
Remember that a function \( f(x) \) is continuous at an accumulation point \( x_0 \) in the function’s domain if and only if the following limit exists and it is finite:
\[ \lim_{x \to x_0} f(x) \]
and if it also equals the evaluation \( f(x_0) \).
Differentiability of a function
Remember that a function\( f(x) \) can be differentiated at a point \( x_0 \) on its domain if and only if the following limit exists and it is finite:
\[ \lim_{\Delta x \to 0} \dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x } \]
If that is the case, the evaluation of the above limit just equals the derivative of the function at the point \( x_0 \).
Connection between continuity and differentiability
Differentiability is a stronger condition than continuity. In other words, if a function can be differentiated at a point, it is also continuous on that very same point. Simply put: the differentiability of a function at a given point implies continuity at that very same point.
Proof
If \( f(x) \) can be differentiated at \( x_0 \), then the following limit exists and it is finite:
\[ \lim_{\Delta x \to 0} \dfrac{f(x_0+\Delta x)- f(x_0)}{\Delta x} \]
For the quotient rule of limits, we may rewrite the above limit as:
\[ \dfrac{\displaystyle \lim_{\Delta x \to 0}f(x_0+\Delta x)- f(x_0) }{\displaystyle \lim_{\Delta x \to 0} \Delta x} \]
We can then say that \( \displaystyle \lim_{\Delta x \to 0} \Delta x = 0 \) by direct substitution. Now, for a theorem from limits theory we have:
\[ \text{if it exists finite} \quad \lim_{x \to x_0} \dfrac{g(x)}{h(x)}, \quad \text{and if} \quad \lim_{x \to x_0} h(x) = 0 \quad \Rightarrow \quad \lim_{x \to x_0} g(x)=0 \]
For a proof, assume by contradiction that \( \displaystyle \lim_{x \to x_0} g(x) = l \neq 0 \). Let’s put:
\[ k(x) = \dfrac{g(x)}{h(x)} \]
As a consequence \( g(x) = h(x) \cdot k(x) \). But since we stated that \( \displaystyle \lim_{x \to x_0} g(x) = l \neq 0 \), then we get:
\[ \lim_{x \to x_0}g(x) = \lim_{x \to x_0} \left(h(x) \cdot k(x) \right) = l \neq 0 \]
But since we assumed \( \displaystyle \lim_{x \to x_0} h(x) = 0 \), for the product rule of limits \( \displaystyle \lim_{x \to x_0} \left(h(x) \cdot k(x) \right) = \lim_{x \to x_0} h(x) \cdot \lim_{x \to x_0} k(x) = 0 \). But this is in contradiction with the assumption at the beginning, hence the proof.
Let’s now use the above theorem to solve our problem. Since the following limit exists:
\[ \lim_{\Delta x \to 0} \dfrac{f(x_0+\Delta x)- f(x_0)}{\Delta x} \]
and since \( \displaystyle \lim_{\Delta x \to 0}\Delta x = 0 \), then we have:
\[ \lim_{\Delta x \to 0}{f(x_0+\Delta x)- f(x_0)} = 0 \qquad (*) \]
Now, we can also define the derivative of a function at a given point as:
\[ f'(x_0) = \lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0} \]
As a consequence expression (*) is the same as:
\[ \lim_{x \to x_0} {f(x)-f(x_0)} = 0 \]
and eventually we go back to the definition of continuity of a function at a given point:
\[ \lim_{x \to x_0} f(x) = f(x_0) \]
That proves that if a function can be differentiated at a point \( x_0 \), then it is also continuous at that point. Just as we wanted. 🙂
Some final thoughts about the connection between differentiability and continuity of a function at a given point
So we have proved the following statement:
“if a function is differentiable at a point, it is also continuous at that very same point”.
This is a statement of the kind \( A \Rightarrow B \). But as a consequence we can also say that:
\[ \neg B \Rightarrow \neg A \]
So, we can also state that if a function is not continuous at a given point, then it also cannot be differentiated at that very same point.
Now one last note. If a function happens to be continuous at a given point, then we cannot be sure that it is also differentiable at that point.
A common example is the function \( y=|x| \), which is continuous at the point \( x = 0 \) but cannot be differentiated at that point.
So we must put a stress on an important fact. The continuity of a function at a given point doesn’t necessarily mean that we can also differentiate the function at that very same point.
Summing up the whole thing:
- if a function is differentiable at a given point, than we can automatically say that the function is also continuous at that point;
- in order to say that a function is differentiable at a given point we must have continuity at that point. But, continuity at a point by itself doesn’t imply differentiability at that very same point.
That’s all about the connection between differentiability and continuity of a function. On the next lesson we will deal with basic derivation rules.