L'Hopital's Rule

We argue for $I = (a,b)$, so the limit is $x \to a^+$; the case $I = (c,a)$ (limit $x \to a^-$) is identical with inequalities reversed.

Step 1: Extend $f,g$ continuously to $a$. Define $f(a) := 0$ and $g(a) := 0$. By hypothesis (2), these definitions make $f$ and $g$ continuous at $a$ from the right, since $\lim_{x\to a^+} f(x) = 0 = f(a)$ and likewise for $g$. On $[a, b) \cap I \cup \{a\}$, $f$ and $g$ are now continuous on $[a, x]$ and differentiable on $(a,x)$ for every $x \in (a,b)$.

Step 2: $g(x) \neq 0$ for $x$ near $a$, $x \neq a$. Suppose for contradiction $g(x_0) = 0$ for some $x_0 \in (a,b)$. Since also $g(a) = 0$, Rolle's Theorem (a special case of the Mean Value Theorem, applied on $[a,x_0]$) gives some $\xi \in (a, x_0)$ with $g'(\xi) = 0$, contradicting hypothesis (1). Hence $g(x) \neq 0$ for all $x \in (a, b)$, and the ratio $f(x)/g(x)$ is defined throughout $(a,b)$.

Step 3: Apply the Cauchy Mean Value Theorem on $[a,x]$. Fix $x \in (a,b)$. On the interval $[a,x]$, $f$ and $g$ are continuous (Step 1) and differentiable on $(a,x)$, and $g'$ never vanishes on $(a,x)$. By the Cauchy Mean Value Theorem, there exists $c_x \in (a,x)$ such that

Since $f(a) = g(a) = 0$ and $g(x) \neq 0$, $g'(c_x) \neq 0$, we may divide to get

This holds for every $x \in (a,b)$; the point $c_x$ depends on $x$ but always lies strictly between $a$ and $x$.

Step 4: Pass to the limit $x \to a^+$. Fix $\varepsilon > 0$. Because $\lim_{t \to a^+} \frac{f'(t)}{g'(t)} = L$, there exists $\delta > 0$ (with $(a, a+\delta) \subset I$) such that

(if $L$ is finite; if $L = +\infty$, replace this with: for every $M$ there is $\delta$ such that $0 < t-a<\delta \Rightarrow f'(t)/g'(t) > M$, and argue symmetrically — see the remark below). Now for any $x$ with $0 < x - a < \delta$, we have $c_x \in (a,x)$, so $0 < c_x - a < x - a < \delta$. By Step 3 and the choice of $\delta$,

Since $\varepsilon > 0$ was arbitrary, this proves $\displaystyle\lim_{x \to a^+} \frac{f(x)}{g(x)} = L$ when $L$ is finite.

Step 5: The case $L = \pm\infty$. Suppose $L = +\infty$. Given $M \in \mathbb{R}$, choose $\delta > 0$ so that $0 < t - a < \delta \implies f'(t)/g'(t) > M$. For $x$ with $0 < x-a < \delta$, Step 3 gives $c_x \in (a,x) \subset (a, a+\delta)$, so

Hence $f(x)/g(x) \to +\infty$ as $x \to a^+$. The case $L = -\infty$ is symmetric (using $<$ instead of $>$). This completes the proof of the one-sided rule. The two-sided statement (limit as $x \to a$, with $I$ a punctured neighborhood of $a$) follows by applying the one-sided result separately from the left and from the right, and noting both one-sided limits equal $L$, hence the two-sided limit is $L$ (using the standard fact that a two-sided limit exists and equals $L$ iff both one-sided limits exist and equal $L$). $\blacksquare$

We prove the case $L$ finite; the infinite cases follow by the same one-sided inequality argument as Step 5 above, applied to the estimate constructed below. Fix $\varepsilon > 0$. By hypothesis (3), choose $\delta_1 > 0$ with $(a, a+\delta_1) \subset I$ such that

Fix a reference point $y := a + \delta_1/2 \in (a, a+\delta_1)$. For any $x \in (a, y)$, apply the Cauchy Mean Value Theorem on $[x,y]$: since $g'$ never vanishes on $I \supset [x,y]$, as in the previous proof $g(x) \neq g(y)$ (else Rolle's theorem would give a zero of $g'$ in $(x,y)$), and there is $c_x \in (x,y) \subset (a, a+\delta_1)$ with

By $(\ast)$, since $c_x \in (a, a+\delta_1)$,

Now algebraically rewrite, for $x$ close enough to $a$ that $g(x) \neq 0$ (possible since $|g(x)|\to\infty$ as $x\to a^+$, so $g(x) \ne 0$ eventually):

— rather than expand this awkwardly, we isolate directly. Write

We redo this cleanly. From

divide both sides by $g(x)$ (nonzero for $x$ near $a$):

Subtract $L$ from both sides and use $(\ast\ast)$-type bound $\frac{f'(c_x)}{g'(c_x)} = L + \eta(x)$ with $|\eta(x)| < \varepsilon/2$ (this is exactly $(\ast)$ since $c_x \in (a,a+\delta_1)$):

Since $y$ (hence $f(y), g(y)$) is now a fixed constant and $|g(x)| \to \infty$ as $x \to a^+$, the quantities $\frac{f(y)}{g(x)} \to 0$ and $\frac{g(y)}{g(x)} \to 0$ as $x \to a^+$. Choose $\delta_2 \in (0, y - a)$ small enough that for $0 < x - a< \delta_2$:

concretely: pick $\delta_2$ so that both $|f(y)/g(x)| < \varepsilon/4$ and $|g(y)/g(x)| < \dfrac{\varepsilon}{4(1+|L|)}$ hold. Then also $|\eta(x)(1 - g(y)/g(x))| \le \frac{\varepsilon}{2}\left(1 + \frac{\varepsilon}{4(1+|L|)}\right) \le \frac{\varepsilon}{2}\cdot 2 = \varepsilon$ once $\varepsilon \le 4(1+|L|)$, which we may assume by shrinking $\varepsilon$ if necessary at the start (limits are unaffected). Combining all three terms with the triangle inequality, for $0 < x-a<\delta_2$:

after absorbing constants (formally: given any $\varepsilon'>0$, run the argument with $\varepsilon = \varepsilon'/3$ scaled appropriately in each of the three error budgets — the mechanism is that each of the three terms is made $<\varepsilon'/3$ independently, which is achievable since two terms vanish as $x\to a^+$ with $y,f(y),g(y),L$ fixed, and the third is controlled by hypothesis (3)). Hence $\displaystyle\lim_{x\to a^+}\frac{f(x)}{g(x)} = L$. $\blacksquare$

Substitute $x = 1/t$, $t \to 0^+$. Define $F(t) = f(1/t)$, $G(t) = g(1/t)$ for $t \in (0, 1/N)$. By the chain rule, $F$ and $G$ are differentiable with

so $G'(t) \ne 0$ (since $g'(1/t) \ne 0$), and

because $1/t \to \infty$ as $t \to 0^+$. Also $F(t) = f(1/t) \to 0$ (resp. $|G(t)|\to\infty$) as $t\to 0^+$ exactly when $f(x)\to 0$ (resp. $|g(x)|\to\infty$) as $x \to \infty$. So $F,G$ satisfy the hypotheses of the finite-$a$ theorem at $a=0^+$, giving

i.e. $\lim_{x\to\infty} f(x)/g(x) = L$, since $F(t)/G(t) = f(1/t)/g(1/t) = f(x)/g(x)$ under $x=1/t$. $\blacksquare$