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 $f(x) = \frac{(x – \xi)^{p}}{p!}f^{(p)}(x_{1}),\quad \phi(x) = \frac{(x – \xi)^{q}}{q!}\phi^{(q)}(x_{2}).$

Example LVIII

1. Find the limit of $\{x – (n + 1)x^{n+1} + nx^{n+2}\}/(1 – x)^{2},$ as $$x \to 1$$. [Here the functions and their first derivatives vanish for $$x = 1$$, and $$f”(1) = n(n + 1)$$, $$\phi”(1) = 2$$.]

2. Find the limits as $$x \to 0$$ of $(\tan x – x)/(x – \sin x),\quad (\tan nx – n\tan x)/(n\sin x – \sin nx).$

3. Find the limit of $$x\{\sqrt{x^{2} + a^{2}} – x\}$$ as $$x \to \infty$$. [Put $$x = 1/y$$.]

4. Prove that $\lim_{x \to n} (x – n)\csc x\pi = \frac{(-1)^{n}}{\pi},\quad \lim_{x \to n} \frac{1}{x – n} \left\{ \csc x\pi – \frac{(-1)^{n}}{(x – n)\pi} \right\} = \frac{(-1)^{n}\pi}{6},$ $$n$$ being any integer; and evaluate the corresponding limits involving $$\cot x\pi$$.

5. Find the limits as $$x \to 0$$ of $\frac{1}{x^{3}}\left(\csc x – \frac{1}{x} – \frac{x}{6}\right),\quad \frac{1}{x^{3}}\left(\cot x – \frac{1}{x} + \frac{x}{3}\right).$

6. $$(\sin x\arcsin x – x^{2})/x^{6} \to \frac{1}{18}$$, $$(\tan x\arctan x – x^{2})/x^{6} \to \frac{2}{9}$$, as $$x \to 0$$.