Proof. For \(x\neq x_0\) we can write \(f(x) - f(x_0) = \frac {f(x) - f(x_0)}{x - x_0}\cdot (x - x_0)\). By the product rule for limits, \[\lim _{x\rightarrow x_0}\bigl (f(x) - f(x_0)\bigr ) = \lim _{x\rightarrow x_0} \frac {f(x) - f(x_0)}{x - x_0}\cdot \lim _{x\rightarrow x_0}(x - x_0) = f'(x_0)\cdot 0 = 0 .\] Hence \(\lim _{x\rightarrow x_0} f(x) = f(x_0)\), and \(f\) is continuous at \(x_0\). (At an end point, use the one-sided limit.)

The converse is false: a continuous function need not be differentiable, as the next example shows. □

Solution. Since \(\bigl ||x| - 0\bigr | = |x|\), \(f(x)\longrightarrow f(0)\) as \(x\longrightarrow 0\), so \(f\) is continuous at \(0\). For the derivative, \(\frac {f(x) - f(0)}{x - 0} = \frac {x}{x}= 1\) when \(x> 0\), and \(\frac {-x}{x} = -1\) when \(x < 0\). Hence \[\lim _{x\rightarrow 0^+} \frac {f(x) - f(0)}{x - 0} = 1 \qquad \text {but}\qquad \lim _{x\rightarrow 0^-}\frac {f(x) - f(0)}{x - 0} = -1 .\] Since \(0\) is an interior point of \(\mathbb {R}\) and the one-sided limits differ, the difference quotient has no limit at \(0\), and \(f\) is not differentiable at the origin.

Proof. By the previous theorem \(f\) and \(g\) are continuous at \(c\), so \(f(x)\longrightarrow f(c)\) and \(g(x)\longrightarrow g(c)\) as \(x\longrightarrow c\). All limits below are as \(x\longrightarrow c\).

(a) \(\displaystyle \frac {(f+g)(x) - (f+g)(c)}{x - c} = \frac {f(x) - f(c)}{x - c} + \frac {g(x) - g(c)}{x - c}\longrightarrow f'(c) + g'(c)\).

(b) \(\displaystyle \frac {af(x) - af(c)}{x - c} = a\,\frac {f(x) - f(c)}{x - c}\longrightarrow af'(c)\).

(c) Adding and subtracting \(g(x)f(c)\), \[\frac {f(x) g(x) - f(c) g(c)}{x - c} = g(x)\,\frac {f(x) - f(c)}{x - c} + f(c)\,\frac {g(x) - g(c)}{x - c}\longrightarrow g(c) f'(c) + f(c) g'(c) .\]

(d) Since \(g(c)\neq 0\) and \(g\) is continuous at \(c\), \(g(x)\neq 0\) near \(c\) (Theorem 2.3.3). For such \(x\), \[\frac {\frac {1}{g(x)} - \frac {1}{g(c)}}{x - c} = -\frac {1}{g(x) g(c)}\cdot \frac {g(x) - g(c)}{x - c}\longrightarrow -\frac {g'(c)}{g(c)^2} .\]

(e) Write \(\frac {f}{g} = f\cdot \frac {1}{g}\) and use (c) and (d): \[\left (\frac {f}{g}\right )'(c) = \frac {f'(c)}{g(c)} - \frac {f(c) g'(c)}{g(c)^2} = \frac {g(c) f'(c) - f(c) g'(c)}{g(c)^2} .\] There are analogous results for one-sided derivatives at end points. □

Proof. Write \(h = g\circ f\). By the remark, there exists a function \(L: S\longrightarrow \mathbb {R}\), continuous at \(c\), such that \begin {equation} \label {eq:chainL} f(x) - f(c) = L(x)\, (x - c) \quad \text {for all } x\in S, \end {equation} and \(L(c) = f'(c)\). Similarly, there is a function \(K: T\longrightarrow \mathbb {R}\), continuous at \(f(c)\), with \(K(f(c)) = g'(f(c))\) and \begin {equation} \label {eq:chainK} g(y) - g(f(c)) = K(y)\, (y - f(c)) \quad \text {for all } y \in T. \end {equation} If \(x\in S\) then \(f(x)\in T\); putting \(y = f(x)\) in (4.2) and using (4.1) gives \[h(x) - h(c) = K(f(x))\, \bigl (f(x) - f(c)\bigr ) = K(f(x))\, L(x)\, (x - c) .\] The function \((K\circ f) L\) is continuous at \(c\), because \(f\) is continuous at \(c\) and \(K\) is continuous at \(f(c)\). By the remark, \(h\) is differentiable at \(c\) and \[h'(c) = K(f(c))\, L(c) = g'(f(c))\, f'(c) .\] □

Solution.

(a)

Both functions are continuous at every \(x\neq 0\), as products and composites of continuous functions. At the origin, \(|f(x) - f(0)| = \left |x \sin \frac {1}{x}\right | \leq |x|\longrightarrow 0\), so \(f\) is continuous at \(0\). Since \(g(x) = x f(x)\), \(g\) is continuous everywhere too.

(b)

For \(x\neq 0\), the chain rule gives \(\frac {d}{dx}\sin \frac {1}{x} = -\frac {1}{x^2} \cos \frac {1}{x}\), and the product rule gives \[f'(x) = \sin \frac {1}{x} - \frac {1}{x} \cos \frac {1}{x}, \qquad g'(x) = 2x \sin \frac {1}{x} - \cos \frac {1}{x} \qquad (x\neq 0).\] At the origin, \[\frac {f(x) - f(0)}{x - 0} = \sin \frac {1}{x},\] which has no limit as \(x\longrightarrow 0\) (see the examples on limits in Chapter 2). So \(f\) is not differentiable at \(0\). On the other hand \[\frac {g(x) - g(0)}{x - 0} = x \sin \frac {1}{x}\longrightarrow 0,\] so \(g\) is differentiable at the origin with \(g'(0) = 0\). Note that \(g'(x)\) has no limit as \(x\longrightarrow 0\), because of the term \(\cos \frac {1}{x}\): \(g\) is differentiable everywhere, but \(g'\) is not continuous at \(0\).

Solution. For \(x^n\), use \(x^n - c^n = (x-c)(x^{n-1} + x^{n-2}c + \dots + c^{n-1})\): \[\frac {x^n - c^n}{x - c} = x^{n-1} + x^{n-2}c + \dots + c^{n-1}\longrightarrow nc^{n-1}\quad (x\rightarrow c),\] since each of the \(n\) terms tends to \(c^{n-1}\). So \(f'(c) = nc^{n-1}\).

For \(\sqrt {x}\) at \(c>0\), multiply by the conjugate: \[\frac {\sqrt {x} - \sqrt {c}}{x - c} = \frac {1}{\sqrt {x}+\sqrt {c}}\longrightarrow \frac {1}{2\sqrt {c}} .\] At \(c = 0\) the one-sided difference quotient \(\frac {\sqrt {x}}{x} = \frac {1}{\sqrt {x}}\longrightarrow \infty \), so \(\sqrt {x}\) is not differentiable at \(0\): its graph has a vertical tangent there.

Solution. For \(x>0\), \(f(x) = x^2\) and \(f'(x) = 2x\); for \(x<0\), \(f(x) = -x^2\) and \(f'(x) = -2x\). At \(0\), \[\frac {f(x) - f(0)}{x - 0} = |x|\longrightarrow 0,\] so \(f'(0) = 0\). In all cases \(f'(x) = 2|x|\). By the example on \(|x|\), \(f'\) is continuous but not differentiable at \(0\). So \(f\) is differentiable once, but not twice, at the origin.

Solution. Let \(u(x) = \frac {x}{1+x^2}\). By the quotient rule, \[u'(x) = \frac {(1+x^2)\cdot 1 - x\cdot 2x}{(1+x^2)^2} = \frac {1-x^2}{(1+x^2)^2},\] and by the chain rule \(h'(x) = \cos (u(x))\,u'(x)\). Since \(|u(x)|\leq \frac {1}{2}<\frac {\pi }{2}\) (because \(1+x^2\geq 2|x|\)), the factor \(\cos u(x)\) is never \(0\). So \(h'(x) = 0\) exactly when \(1 - x^2 = 0\), that is, at \(x = \pm 1\).