Proof. Since \(g^{(n)}\) is continuous on \((a,b)\) and \(g^{(n)} \neq 0\) there is some \(r> 0\) with \((\alpha - r, \alpha + r)\subseteq (a,b)\) and such that \(g^{(n)}(t)\neq 0\) for all \(t\) such that \(\alpha - r < t < \alpha + r\). Let \(x\) be any real number such that \(0 < \begin {vmatrix} x - \alpha \\ \end {vmatrix} < r\), then \(f, \, g\) and their first \(n\) derivatives are continuous on the closed interval \([\alpha , x]\). By Taylor’s Theorem \[f(x) = f(\alpha ) + f'(\alpha )\, (x - \alpha ) + \cdots \cdots \cdots + \dfrac {f^{(n - 1)}(\alpha ) \, (x - \alpha )^{(n - 1)}}{(n - 1)!} + \dfrac {f^n(c)\, (x-\alpha )^n}{n!}\] \(\displaystyle {\implies \hspace {0.3cm} f(x) = \dfrac {f^n (c)\, (x - \alpha )^n}{n!}}\hspace {0.3cm}\) where \(\, \alpha < c < x\).

Similarly \[g(x) = g(\alpha ) + g'(\alpha )\, (x - \alpha ) + \cdots \cdots \cdots + \dfrac {g^{(n - 1)}(\alpha ) \, (x - \alpha )^{(n - 1)}}{(n - 1)!} + \dfrac {g^n(d)\, (x-\alpha )^n}{n!}\] \(\displaystyle {\implies \hspace {0.3cm} g(x) = \dfrac {g^n (d)\, (x - \alpha )^n}{n!}}\hspace {0.3cm}\) where \(\, \alpha < d < x\).

Thus \(\, 0 < \begin {vmatrix} \alpha - d\\ \end {vmatrix} < r\) since \(\, 0 < \begin {vmatrix} \alpha - x\\ \end {vmatrix} < r\). Since \(g^{(n)}(t) \neq 0\) for all \(t\) such that \(\alpha - r < t < \alpha + r\), we see that \(g^n(d)\neq 0\) and so \(g(x) \neq 0\) for \(\, 0<\begin {vmatrix} x - \alpha \\ \end {vmatrix} < r\) \(\, (x\neq \alpha \) but \(x\) is in a small neighbourhood of \(\alpha )\). Therefore, if \(0< \begin {vmatrix} x - \alpha \\ \end {vmatrix} < r\), then \(\, \dfrac {f(x)}{g(x)} = \dfrac {f^n(c)}{g^n(c)}\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm}(*)\). Now as \(x\longrightarrow \alpha \) we have \(c\longrightarrow \alpha \) and \(\, d\longrightarrow \alpha \) because \(c\) and \(d\) are between \(\alpha \) and \(x\).
Further \(f^{(n)}(c)\longrightarrow f^{(n)}(\alpha )\) as \(c\longrightarrow \alpha \) and \(g^{(n)}(c)\longrightarrow ^{(n)}(\alpha )\) and \(d\longrightarrow \alpha \) because \(f^{(n)}\) and \(g^{(n)}\) are continuous. Thus \[\lim _{c\,, d\rightarrow \alpha } \, \dfrac {f^{(n)}(c)}{g^{(n)}(d)} = \dfrac {f^n(\alpha )}{g^n(\alpha )}\] Hence, from \((*)\) we have \(\hspace {0.2cm}\displaystyle {\lim _{x\rightarrow \alpha }\, \dfrac {f(x)}{g(x)} = \dfrac {f^n(\alpha )}{g^n(\alpha )}}\).

Note that Taylor’s Theorem states that a number \(c\) exists with given properties when \(f\) satisfies certain conditions. The number \(c\) is determined by the function \(f\) and the interval \([a,b]\) and it is not just any arbitrary number between \(a\) and \(b\). Hence the use of \(c\) and \(d\) in the above proof.

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Solution. Let \(f(x) = \cos x - 1\, , \hspace {0.2cm} g(x) = x^2\,\) for all real values of \(x\). Then \(f(0) = 0\) and \(g(0) = 0\).
We must now check that all conditions for the application of L’H\(\hat {\text {o}}\)pital’s Rule are satisfied before applying it. In particular we have to determine what \(n\) we must use differentiating \(g\) at \(x = 0\). \begin {align*} g'(x) & = 2x \hspace {0.3cm}\implies \hspace {0.3cm} g'(0) = 0\\\\ g''(x) & = 2 \hspace {0.3cm}\implies \hspace {0.3cm} g''(0) = 2 \end {align*}

Therefore \(n = 2\), we have that \begin {align*} f'(x) & = -\sin x\hspace {0.3cm}\implies \hspace {0.3cm} f'(0) = 0\\\\ f''(x) & = -\cos x\hspace {0.3cm}\implies \hspace {0.3cm} f''(0) = 1 \end {align*}

We see that \(f,\, g\) and all their derivatives are continuous on \(\mathbb {R}\). Hence using L’H\(\hat {\text {o}}\)pital’s Rule \[\lim _{x\longrightarrow 0}\, \dfrac {\cos x - 1}{x^2} = \dfrac {f''(0)}{g''(0)} = \dfrac {-1}{2}\]