**B. The calculation of certain limits.** Suppose that \(f(x)\) and \(\phi(x)\) are two functions of \(x\) whose derivatives \(f'(x)\) and \(\phi'(x)\) are continuous for \(x = \xi\) and that \(f(\xi)\) and \(\phi(\xi)\) are both equal to zero. Then the function \[\psi(x) = f(x)/\phi(x)\] is not defined when \(x = \xi\). But of course it may well tend to a limit as \(x \to \xi\).

Now \[f(x) = f(x) – f(\xi) = (x – \xi)f'(x_{1}),\] where \(x_{1}\) lies between \(\xi\) and \(x\); and similarly \(\phi(x) = (x – \xi)\phi'(x_{2})\), where \(x_{2}\) also lies between \(\xi\) and \(x\). Thus \[\psi(x) = f'(x_{1})/\phi'(x_{2}).\] We must now distinguish four cases.

(1) If neither \(f'(\xi)\) nor \(\phi'(\xi)\) is zero, then \[f(x)/\phi(x) \to f'(\xi)/\phi'(\xi).\]

(2) If \(f'(\xi) = 0\), \(\phi'(\xi) \neq 0\), then \[f(x)/\phi(x) \to 0.\]

(3) If \(f'(\xi) \neq 0\), \(\phi'(\xi)= 0\), then \(f(x)/\phi(x)\) becomes numerically very large as \(x \to \xi\): but whether \(f(x)/\phi(x)\) tends to \(\infty\) or \(-\infty\), or is sometimes large and positive and sometimes large and negative, we cannot say, without further information as to the way in which \(\phi'(x) \to 0\) as \(x \to \xi\).

(4) If \(f'(\xi) = 0\), \(\phi'(\xi) = 0\), then we can as yet say nothing about the behaviour of \(f(x)/\phi(x)\) as \(x \to 0\).

But in either of the last two cases it may happen that \(f(x)\) and \(\phi(x)\) have continuous second derivatives. And then \[\begin{aligned} f(x) &= f(x) – f(\xi) – (x – \xi)f'(\xi) = \tfrac{1}{2}(x – \xi)^{2} f”(x_{1}),\\ \phi(x) &= \phi(x) – \phi(\xi) – (x – \xi)\phi'(\xi) = \tfrac{1}{2}(x – \xi)^{2} \phi”(x_{2}),\end{aligned}\] where again \(x_{1}\) and \(x_{2}\) lie between \(\xi\) and \(x\); so that \[\psi(x)= f”(x_{1})/\phi”(x_{2}).\] We can now distinguish a variety of cases similar to those considered above. In particular, if neither second derivative vanishes for \(x = \xi\), we have \[f(x)/\phi(x) \to f”(\xi)/\phi”(\xi).\]

It is obvious that this argument can be repeated indefinitely, and we obtain the following theorem:

*suppose that \(f(x)\) and \(\phi(x)\) and their derivatives, so far as may be wanted, are continuous for \(x = \xi\). Suppose further that \(f^{(p)}(x)\) and \(\phi^{(q)}(x)\) are the first derivatives of \(f(x)\) and \(\phi(x)\) which do not vanish when \(x = \xi\). Then*

*(1) if \(p = q\), \(f(x)/\phi(x) \to f^{(p)}(\xi)/\phi^{(p)}(\xi)\);*

*(2) if \(p > q\), \(f(x)/\phi(x) \to 0\);*

*(3) if \(p < q\), and \(q – p\) is even, either \(f(x)/\phi(x) \to +\infty\) or \(f(x)/\phi(x) \to -\infty\), the sign being the same as that of \(f^{(p)}(\xi)/\phi^{(q)}(\xi)\);*

*(4) if \(p < q\) and \(q – p\) is odd, either \(f(x)/\phi(x) \to +\infty\) or \(f(x)/\phi(x) \to -\infty\), as \(x \to \xi+0\), the sign being the same as that of \(f^{(p)}(\xi)/\phi^{(q)}(\xi)\), while if \(x \to \xi – 0\) the sign must be reversed.*

This theorem is in fact an immediate corollary from the equations \[f(x) = \frac{(x – \xi)^{p}}{p!}f^{(p)}(x_{1}),\quad \phi(x) = \frac{(x – \xi)^{q}}{q!}\phi^{(q)}(x_{2}).\]

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