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Mathematics LibreTexts

13.5: Continuity and Derivatives

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    24881
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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:

    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 \(x = c\).

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