13.5: Continuity and Derivatives

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Contributed by Richard Hammack
Professor (Mathematics) at Virginia Commonwealth University

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A major purpose of limits is that they can give information about how a function behaves near a “bad point” $x = c$. Even if $f(c)$ is not defined, it may be that $\lim_{x \rightarrow c} f(x) = L$, for some number L. In this event we know that $f(x)$ becomes ever closer to L as x approaches the forbidden c.

Of course not every value $x = c$ is a “bad point.” It could be that $f(c)$ is defined, and, moreover, $\lim_{x \rightarrow c} f(x) = f(c)$. If this is the case for every c in the domain of $f(x)$, then we say that f is continuous. Issues concerning whether or not f is continuous are called issues of continuity.

In a first course in calculus it is easy to overlook the huge importance of continuity. And happily, we can (in a first course) almost ignore it. But in fact, the theoretical foundation of calculus rests on continuity. Roughly speaking, there are countless theorems having the form

If f is continuous, then f has some significant property.

Continuity allows us to draw certain important conclusions about a function. Here is its definition.

Definition 13.3

A function f (x) is continuous at $x = c$ if $\lim_{x \rightarrow c} f(x) = f(c)$.

Note that this means all of the following three conditions must be met:

$f(c)$ is defined,
$\lim_{x \rightarrow c} f(x)$ exists,

3. $\lim_{x \rightarrow c} f(x) = f(c)$.

If one or more of these conditions fail, then f (x) is discontinuous at c.

To illustrate this definition, five functions $f(x)$ are graphed below. Only the function on the far right is continuous at $x = c$.

$Screen Shot 2020-06-23 at 4.18.30 PM.png$

Most familiar functions are continuous at each point $x = c$ in their domain. For instance, exercises 3 and 4 in the previous section imply that polynomials and rational functions are continuous at any number c in their domains. One application of continuity is a limit law for composition. The previous section might prompt us to conjecture that $\lim_{x \rightarrow c} f(g(x)) = f(\lim_{x \rightarrow c} g(x))$. However, this does not hold without an assumption of continuity.

Theorem 13.9 (Composition rule)

If $\lim_{x \rightarrow c} g(x)$ and f is continuous at $x = L$, then $\lim_{x \rightarrow c} f(g(x)) = f(\lim_{x \rightarrow c} g(x))$.

Proof. Suppose $\lim_{x \rightarrow c} g(x)$ and f is continuous at $x = L$. We need to show $\lim_{x \rightarrow c} f(g(x)) = f(\lim_{x \rightarrow c} g(x))$. According to Definition 13.2, for any $epsilon > 0$ we must show there is a corresponding $\delta > 0$ for which $0 < |x-c| < \delta$ implies $|f(g(x))-L| < \epsilon$.

So let $\epsilon > 0$. As f is continuous at L, Definition 13.3 yields \lim_{x \rightarrow L} f(x) = f(L)\).

From this, we know there is a real number $\delta' > 0$ for which $|x-L|< \delta'$ implies $|f(x)-f(L)| < \epsilon$. (A)

But also, from $\lim_{x \rightarrow c} g(x) = L$, we know that there is a real number $\delta > 0$ for which $0 < |x-c| < \delta$ implies $|g(x)-L| < \delta'$.

If $0 < |x-c| < \delta$, then we have $|g(x)-L| < \delta'$, and from this (A) yields

$|f(g(x))-f(L)| < \epsilon$. Thus $\lim_{x \rightarrow c} f(g(x)) = f(L)$, and the proof is complete.

In calculus you learned that the derivative of a real-valued function f is another function f′ for which $f′(c)$ is defined as

$f′(c) = \lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$,

provided the limit exists (in which case we say f is differentiable at c). You may recall that differentiability implies continuity.

Theorem 13.10

If f is differentiable at c, then f is continuous at c.

Proof. Suppose f is differentiable at c, so $\lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c} = f'(c)$. Write $f(x)$ as

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

Taking limits of both sides and using limit laws,

$\lim_{x \rightarrow c} f(x)= (\lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}) \cdot \lim_{x \rightarrow c} (x-c) + \lim_{x \rightarrow c} f(c) = f (c) \cdot 0+f(c) = f(c)$.

Thus $\lim_{x \rightarrow c} f(x) = f(c)$, which means f is continuous at c.

Exercise

Exercise $\PageIndex{1}$

Prove that the function $f(x) = \sqrt{x}$ is continuous at any number $c > 0$. Deduce that $\lim_{x \rightarrow c} \sqrt{g(x)} = \sqrt{\lim_{x \rightarrow c} g(x)}$, provided $\lim_{x \rightarrow c} g(x)$ exists and is greater than zero.

Exercise $\PageIndex{2}$

Show that the condition of continuity in Theorem 13.9 is necessary by finding functions f and g for which $\lim_{x \rightarrow c} g(x) = L$, and f is not continuous at $x = L$, and $\lim_{x \rightarrow c} f(g(x)) \ne f(\lim_{x \rightarrow c} g(x))$.