Proof. Since \(g^{(n)}\) is continuous and \(g^{(n)}(\alpha ) \neq 0\), there is some \(r> 0\) with \((\alpha - r, \alpha + r)\subseteq (a,b)\) and \(g^{(n)}(t)\neq 0\) for all \(t\in (\alpha - r, \alpha + r)\). Let \(0 < |x - \alpha | < r\). By Taylor’s Theorem, using (i) and (ii), there are points \(c\) and \(d\) strictly between \(\alpha \) and \(x\) such that \[f(x) = \frac {f^{(n)}(c) (x - \alpha )^n}{n!}, \qquad g(x) = \frac {g^{(n)}(d) (x - \alpha )^n}{n!} .\] Since \(|d-\alpha |<r\), \(g^{(n)}(d)\neq 0\), and so \(g(x)\neq 0\). Therefore, for \(0< |x - \alpha | < r\), \begin {equation} \label {eq:lhratio} \frac {f(x)}{g(x)} = \frac {f^{(n)}(c)}{g^{(n)}(d)} . \end {equation} As \(x\longrightarrow \alpha \), also \(c\longrightarrow \alpha \) and \(d\longrightarrow \alpha \), because \(c\) and \(d\) lie between \(\alpha \) and \(x\). By the continuity of \(f^{(n)}\) and \(g^{(n)}\), \(f^{(n)}(c)\longrightarrow f^{(n)}(\alpha )\) and \(g^{(n)}(d)\longrightarrow g^{(n)}(\alpha )\neq 0\). Hence, from (4.10), \(\lim _{x\rightarrow \alpha } \frac {f(x)}{g(x)} = \frac {f^{(n)}(\alpha )}{g^{(n)}(\alpha )}\).

Note that the points \(c\) and \(d\) given by Taylor’s Theorem depend on the function as well as on \(x\), so in general \(c\neq d\); that is why two different letters are needed. □

Solution. (a) Let \(f(x) = \cos x - 1\) and \(g(x) = x^2\); all their derivatives are continuous on \(\mathbb {R}\). We first find the order \(n\) at which \(g\) stops vanishing at \(0\): \[g(0) = 0,\qquad g'(0) = 0,\qquad g''(0) = 2\neq 0,\] so \(n = 2\). Check that \(f\) vanishes to the same order: \(f(0) = 0\), \(f'(x) = -\sin x\) gives \(f'(0) = 0\), and \(f''(x) = -\cos x\) gives \(f''(0) = -1\). Hence \[\lim _{x\rightarrow 0} \frac {\cos x - 1}{x^2} = \frac {f''(0)}{g''(0)} = -\frac {1}{2} .\]

(b) Let \(f(x) = e^{x^2} - 1\) and \(g(x) = \sin x^2\). Then \begin {align*} f'(x) &= 2x e^{x^2}, & f''(x) &= (2+4x^2)e^{x^2},\\ g'(x) &= 2x\cos x^2, & g''(x) &= 2\cos x^2 - 4x^2\sin x^2 , \end {align*}

so \(f(0) = f'(0) = 0 = g(0) = g'(0)\), and \(f''(0) = 2\), \(g''(0) = 2\neq 0\). Hence the limit is \(\frac {2}{2} = 1\).

Solution. (a) With \(f(x) = x - \sin x\) and \(g(x) = x^3\): \(g(0) = g'(0) = g''(0) = 0\) and \(g'''(0) = 6\), so \(n = 3\). Also \(f(0) = 0\), \(f'(x) = 1 - \cos x\), \(f''(x) = \sin x\) and \(f'''(x) = \cos x\), which give \(f'(0) = f''(0) = 0\) and \(f'''(0) = 1\). The limit is \(\frac {1}{6}\).

(b) With \(f(x) = e^x - 1 - x\) and \(g(x) = x^2\): \(n = 2\), \(f'(x) = e^x - 1\) and \(f''(x) = e^x\), so \(f(0) = f'(0) = 0\) and \(f''(0) = 1\). The limit is \(\frac {1}{2}\).

(c) With \(f(x) = 1 - \cos x\) and \(g(x) = x\sin x\): \(g'(x) = \sin x + x\cos x\) and \(g''(x) = 2\cos x - x\sin x\), so \(g(0) = g'(0) = 0\) and \(g''(0) = 2\). Also \(f'(0) = \sin 0 = 0\) and \(f''(0) = \cos 0 = 1\). The limit is \(\frac {1}{2}\).

Solution. Let \(f(x) = x^2\) and \(g(x) = \sin x\). Here \(g(0) = 0\) but \(g'(0) = 1\neq 0\), so \(n = 1\). We need \(f(0) = 0\), which holds, and the rule gives \(\frac {f'(0)}{g'(0)} = \frac {0}{1} = 0\). It would be wrong to differentiate a second time: \(\frac {f''(0)}{g''(0)} = \frac {2}{0}\) is meaningless. The number of differentiations is fixed by the first derivative of \(g\) that does not vanish at \(\alpha \), and \(f\) and its earlier derivatives must vanish there.