
# 13.5: Continuity and Derivatives


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:

1. $$f(c)$$ is defined,
2. $$\lim_{x \rightarrow c} f(x)$$ 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 $$x = c$$.

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 $$\lim_{x \rightarrow c} f(g(x)) \ne f(\lim_{x \rightarrow c} g(x))$$.