1. 미분 (iii): 로피탈 법칙과 테일러 정리Differentiation (iii): L’Hospital’s Rule and Taylor’s Theorem

For $c \in [a, b)$, we define

or $f(x) \to -\infty$ as $x \downarrow c$ iff for any $M \lt 0$, there exists a $\delta > 0$ such that

for all $y \in (c, c + \delta)$.

Similarly, for $c \in (a, b]$, we define

or $f(x) \to \infty$ as $x \uparrow c$ iff for any $M > 0$, there exists a $\delta > 0$ such that

for all $y \in (c - \delta, c)$.

Finally, for $c \in (a, b]$, we define

or $f(x) \to -\infty$ as $x \uparrow c$ iff for any $M \lt 0$, there exists a $\delta > 0$ such that

for all $y \in (c - \delta, c)$.

• (i) ($\Leftarrow$) Suppose the condition holds for all $M \in \mathbb{R}$; in particular it holds for all $M > 0$, which is exactly the condition for the right limit in Definition 1.61. Hence $\lim_{x \to c^+} f(x) = \infty$.

  ($\Rightarrow$) Suppose $\lim_{x \to c^+} f(x) = \infty$. Applying Definition 1.61 with $M' = 1 > 0$, there exists a $\delta > 0$ such that

  $$f(y) > 1 \qquad \forall y \in (c, c + \delta).$$

  Now let $M \in \mathbb{R}$ be arbitrary. If $M \leq 0$, then since $M \leq 0 \lt 1$, we have $f(y) > 1 > M$ for all $y \in (c, c+\delta)$, that is, the $\delta$ above works for this $M$ as well. If $M > 0$, then applying the assumption $\lim_{x \to c^+} f(x) = \infty$ directly to this $M$ in Definition 1.61, we obtain a $\delta > 0$ such that $f(y) > M$. In either case, the desired $\delta$ exists, so for all $M \in \mathbb{R}$, there exists a $\delta > 0$ such that $f(y) > M$. This proves (i).

• (ii) ($\Rightarrow$) Suppose $\lim_{x \to c^+} f(x) = \infty$ and let $M_0 := 0$. Then for any $M > M_0 = 0$, by Definition 1.61 there exists a $\delta > 0$ such that $f(y) > M$.

  ($\Leftarrow$) Suppose there exists $M_0 \in \mathbb{R}$ such that for all $M > M_0$, one can find a $\delta > 0$ such that $f(y) > M$. Let $M' > 0$ be given. Let $M := \max\{M', M_0 + 1\}$. Then $M \geq M_0 + 1 > M_0$, so by the assumption there exists a $\delta > 0$ such that

  $$f(y) > M \qquad \forall y \in (c, c+\delta).$$

  Since $M \geq M'$, we have $f(y) > M \geq M'$, that is, $f(y) > M'$ for all $y \in (c, c+\delta)$. Hence this $\delta$ satisfies the condition of Definition 1.61 for $M'$, so $\lim_{x \to c^+} f(x) = \infty$. This proves (ii).

Let $x \in (a, b)$ be arbitrary. Since $\lim_{t \to a^+} f(t) = \infty$, applying Exercise 1.63(i) at the left endpoint $a$, for any $M \in \mathbb{R}$ there exists a $\delta > 0$ such that (Exercise 1.63 is stated for a point of $(a,b)$, but its proof uses only the right-limit condition of Definition 1.61, and Definition 1.61 itself covers $c \in [a,b)$, i.e. includes the endpoint $a$, so the equivalence in (i) holds equally well at the endpoint $a$).

Take $M := f(x)$. Then there exists a $\delta > 0$ such that

If necessary, we may shrink $\delta$ so that $a + \delta \lt b$ (shrinking $\delta$ only makes $(a, a+\delta)$ a smaller interval, so the above inequality still holds). Let $c := a + \delta$. Then $c \in (a, b)$ and

holds. Since $x \in (a,b)$ was arbitrary, the proof is complete.

We consider the case $a = -\infty$. By Remark 1.64, a neighborhood of $-\infty$ has the form $(-\infty, c)$ for $c \in \mathbb{R}$, so, following the pattern of Definition 1.6(ii), for $A \in \mathbb{R}$ we define

to mean that for any $\varepsilon > 0$, there exists $c \in \mathbb{R}$ such that

Under this definition, we follow the proof of Lemma 1.65 verbatim, replacing $(a, a+\delta)$ by $(-\infty, c)$ and omitting the step of "taking $\delta = c - a$".

First we prove "the if part ($\Leftarrow$)". Let $\varepsilon > 0$ be given. Choose $q_1$ and $q_2$ so that

Then by the assumption, there exists a $c \in (a, b) = (-\infty, b)$ such that

Thus

Combining these, we have

Since we already obtained the desired form on $(-\infty, c)$ (there is no need for the step of converting to $\delta = c - a$ as in the finite-$a$ case), $|f(y) - A| \lt \varepsilon$ holds for all $y \in (-\infty, c)$.

Next we prove "the only if part ($\Rightarrow$)". Let $q_1 \lt A \lt q_2$. Define

Then obviously $\varepsilon > 0$ and for this $\varepsilon$, there exists $c_0 \in \mathbb{R}$ such that

Choose $c \in (-\infty, c_0) \cap (a, b)$ (this is nonempty since both $(-\infty, c_0)$ and $(a, b) = (-\infty, b)$ are open intervals extending toward $-\infty$). Then

This inequality implies

and

for all $y \in (-\infty, c)$. This proves the lemma for the case $a = -\infty$.

Assume $-\infty \lt A \leq \infty$. For any real number $q_1$ such that $q_1 \lt A$, we show that there exists a $c \in (a, b)$ such that

Let $q_1 \lt A$ and choose $r \in (q_1, A)$. By (1.20), there exists a $c_0 \in (a, b)$ such that

Let $a \lt x \lt y \lt c_0$. Then by Cauchy's mean-value theorem, there exists a $t_0 \in (a, c_0)$ such that

Suppose (1.21) holds. Then taking $x \downarrow a$ in the above inequality, we have

Next suppose (1.22) holds. Fix $y$ and take $x$ close enough to $a$ so that

that is, there exists $c_1 \in (a, y)$ such that this holds for all $x \in (a, c_1)$. For these $x$, since $\frac{g(x) - g(y)}{g(x)} > 0$, multiplying the above inequality by $\frac{g(x) - g(y)}{g(x)}$ (which preserves the direction of the inequality), we have

and thus

Since (1.22) holds, if $x$ is close enough to $a$, then

that is, there exists $c_2 \in (a, c_1) \subset (a, b)$ such that

for all $x \in (a, c_2)$.

In either case, the desired $c$ exists ($c = c_0$ in the case of (1.21), $c = c_2$ in the case of (1.22)), which completes the proof of step (ii).