[proofplan] We treat the case $L \in \mathbb{R}$ (finite). The idea is to decompose $f(x)/g(x)$ into a Cauchy-quotient term, which [Cauchy's Mean Value Theorem](/theorems/187) controls, and a remainder term, which the divergence $|g(x)| \to \infty$ drives to zero. More precisely: fix a reference point $y$ in the interval where $f'/g'$ is $\epsilon$-close to $L$. Rewrite $f(x)/g(x)$ by adding and subtracting $f(y)$ in the numerator, factor out $g(x) - g(y)$ to expose a Cauchy quotient, then let $x \to a^+$ so that $|g(x)| \to \infty$ absorbs all terms involving the fixed values $f(y)$ and $g(y)$. [/proofplan] [step:Choose $\delta$ so that $f'/g'$ is uniformly $\epsilon$-close to $L$] We prove the case $L \in \mathbb{R}$. Let $\epsilon > 0$. Since $\lim_{x \to a^+} f'(x)/g'(x) = L$, there exists $\delta > 0$ such that $a + \delta < b$ and for every $x \in (a, a + \delta)$, \begin{align*} \left|\frac{f'(x)}{g'(x)} - L\right| < \epsilon. \end{align*} [guided] We prove the case $L \in \mathbb{R}$. The cases $L = +\infty$ and $L = -\infty$ follow by applying the finite case to the reciprocal $g/f$ or by a direct adaptation of the argument below with one-sided bounds replacing $\epsilon$-balls. Let $\epsilon > 0$. The hypothesis $\lim_{x \to a^+} f'(x)/g'(x) = L$ means that for every $\epsilon > 0$ there exists $\delta > 0$ with $a + \delta < b$ such that for all $x \in (a, a + \delta)$, \begin{align*} \left|\frac{f'(x)}{g'(x)} - L\right| < \epsilon. \end{align*} The point of this step is to lock in a neighbourhood $(a, a + \delta)$ inside which the derivative ratio is uniformly controlled. Everything that follows takes place inside this neighbourhood. [/guided] [/step] [step:Fix a reference point $y$ and apply Cauchy's Mean Value Theorem on $[x, y]$] Fix any $y \in (a, a + \delta)$. For each $x \in (a, y)$, the functions $f$ and $g$ are differentiable on $(x, y)$ and continuous on $[x, y]$ (differentiability on an open interval implies continuity there). Since $g'$ is nonzero on $(a, b) \supseteq (x, y)$, [Cauchy's Mean Value Theorem](/theorems/187) applies on $[x, y]$: there exists $c_x \in (x, y) \subset (a, a + \delta)$ with \begin{align*} \frac{f(y) - f(x)}{g(y) - g(x)} = \frac{f'(c_x)}{g'(c_x)}. \end{align*} The denominator $g(y) - g(x)$ is nonzero for all $x$ sufficiently close to $a$: indeed, $g(y)$ is a fixed finite value while $|g(x)| \to \infty$ as $x \to a^+$, so $g(x) \neq g(y)$ once $|g(x)| > |g(y)|$. Since $c_x \in (a, a + \delta)$, the estimate from the previous step gives \begin{align*} \left|\frac{f(y) - f(x)}{g(y) - g(x)} - L\right| = \left|\frac{f'(c_x)}{g'(c_x)} - L\right| < \epsilon. \end{align*} [guided] Fix any $y \in (a, a + \delta)$. The role of $y$ is to serve as a fixed reference point: its values $f(y)$ and $g(y)$ are constants that we will later absorb using $|g(x)| \to \infty$. For each $x \in (a, y)$, we wish to apply [Cauchy's Mean Value Theorem](/theorems/187) on the interval $[x, y]$. We verify the hypotheses: 1. **Continuity on $[x, y]$:** Both $f$ and $g$ are differentiable on the open interval $(a, b) \supseteq [x, y]$, which implies continuity on $[x, y]$. 2. **Differentiability on $(x, y)$:** This follows from differentiability on $(a, b)$. 3. **$g'$ nonzero on $(x, y)$:** Guaranteed by the hypothesis $g'(x) \neq 0$ for all $x \in (a, b)$. All three conditions are met, so there exists $c_x \in (x, y)$ satisfying \begin{align*} \frac{f(y) - f(x)}{g(y) - g(x)} = \frac{f'(c_x)}{g'(c_x)}. \end{align*} We need the denominator $g(y) - g(x)$ to be nonzero. Since $|g(x)| \to \infty$ as $x \to a^+$ while $g(y)$ is a fixed real number, there exists $\delta_1 \in (0, y - a)$ such that $|g(x)| > |g(y)|$ for all $x \in (a, a + \delta_1)$, which forces $g(x) \neq g(y)$. (In fact, [Cauchy's Mean Value Theorem](/theorems/187) itself guarantees $g(y) - g(x) \neq 0$ under the hypothesis $g' \neq 0$, since $g(y) = g(x)$ together with [Rolle's theorem](/theorems/185) would produce a zero of $g'$.) Since $c_x \in (x, y) \subset (a, a + \delta)$, the $\epsilon$-control from the previous step transfers directly: \begin{align*} \left|\frac{f(y) - f(x)}{g(y) - g(x)} - L\right| = \left|\frac{f'(c_x)}{g'(c_x)} - L\right| < \epsilon. \end{align*} This is the key estimate: the Cauchy quotient $(f(y) - f(x))/(g(y) - g(x))$ inherits the $\epsilon$-closeness to $L$ that we established for $f'/g'$. [/guided] [/step] [step:Decompose $f(x)/g(x)$ into the Cauchy quotient and a vanishing remainder] For $x$ sufficiently close to $a$, $g(x) \neq 0$ (since $|g(x)| \to \infty$). Adding and subtracting $f(y)$ in the numerator and factoring: \begin{align*} \frac{f(x)}{g(x)} &= \frac{f(x) - f(y)}{g(x)} + \frac{f(y)}{g(x)} \\ &= \frac{f(x) - f(y)}{g(x) - g(y)} \cdot \frac{g(x) - g(y)}{g(x)} + \frac{f(y)}{g(x)} \\ &= \frac{f(y) - f(x)}{g(y) - g(x)} \cdot \left(1 - \frac{g(y)}{g(x)}\right) + \frac{f(y)}{g(x)}. \end{align*} In the last equality, the sign reversal in both numerator and denominator of the first fraction leaves the Cauchy quotient unchanged. Subtracting $L$ from both sides and collecting: \begin{align*} \frac{f(x)}{g(x)} - L &= \left(\frac{f(y) - f(x)}{g(y) - g(x)} - L\right)\left(1 - \frac{g(y)}{g(x)}\right) + L\left(1 - \frac{g(y)}{g(x)}\right) - L + \frac{f(y)}{g(x)}. \end{align*} The terms $L(1 - g(y)/g(x)) - L = -Lg(y)/g(x)$ combine with $f(y)/g(x)$ to give \begin{align*} \frac{f(x)}{g(x)} - L = \underbrace{\left(\frac{f(y) - f(x)}{g(y) - g(x)} - L\right)}_{\text{(I)}} \cdot \underbrace{\left(1 - \frac{g(y)}{g(x)}\right)}_{\text{(II)}} + \underbrace{\frac{f(y) - Lg(y)}{g(x)}}_{\text{(III)}}. \end{align*} [guided] The goal is to express $f(x)/g(x)$ in a form where every factor is either controlled by the Cauchy estimate or driven to zero by $|g(x)| \to \infty$. Why not just bound $f(x)/g(x)$ directly? Because we have no pointwise control over $f(x)$ or $g(x)$ individually — only the derivative ratio $f'/g'$ is close to $L$. The Cauchy quotient is the bridge between the derivative ratio and the function ratio. For $x$ sufficiently close to $a$, $g(x) \neq 0$ (since $|g(x)| \to \infty$), so the ratio $f(x)/g(x)$ is well-defined. Write \begin{align*} \frac{f(x)}{g(x)} = \frac{f(x) - f(y) + f(y)}{g(x)} = \frac{f(x) - f(y)}{g(x)} + \frac{f(y)}{g(x)}. \end{align*} The second term $f(y)/g(x)$ involves only the fixed constant $f(y)$ divided by $g(x) \to \pm\infty$, so it vanishes as $x \to a^+$. For the first term, we want to introduce the Cauchy quotient. Factor out $(g(x) - g(y))/g(x)$: \begin{align*} \frac{f(x) - f(y)}{g(x)} = \frac{f(x) - f(y)}{g(x) - g(y)} \cdot \frac{g(x) - g(y)}{g(x)}. \end{align*} Reversing the sign in both numerator and denominator of the first factor (which does not change its value) and rewriting the second factor: \begin{align*} \frac{f(x)}{g(x)} = \frac{f(y) - f(x)}{g(y) - g(x)} \cdot \left(1 - \frac{g(y)}{g(x)}\right) + \frac{f(y)}{g(x)}. \end{align*} Now subtract $L$ from both sides. To isolate the Cauchy-quotient error, expand: \begin{align*} \frac{f(x)}{g(x)} - L &= \left(\frac{f(y) - f(x)}{g(y) - g(x)} - L\right)\left(1 - \frac{g(y)}{g(x)}\right) + L\left(1 - \frac{g(y)}{g(x)}\right) - L + \frac{f(y)}{g(x)}. \end{align*} Simplify the last three terms: $L(1 - g(y)/g(x)) - L = -Lg(y)/g(x)$, so $-Lg(y)/g(x) + f(y)/g(x) = (f(y) - Lg(y))/g(x)$. The result is \begin{align*} \frac{f(x)}{g(x)} - L = \underbrace{\left(\frac{f(y) - f(x)}{g(y) - g(x)} - L\right)}_{\text{(I)}} \cdot \underbrace{\left(1 - \frac{g(y)}{g(x)}\right)}_{\text{(II)}} + \underbrace{\frac{f(y) - Lg(y)}{g(x)}}_{\text{(III)}}. \end{align*} The three labelled pieces play distinct roles: - **(I)** is the Cauchy-quotient error, bounded by $\epsilon$ from the previous step. - **(II)** is a multiplicative correction that tends to $1$ as $|g(x)| \to \infty$ (since $g(y)/g(x) \to 0$); it is bounded by $2$ for $x$ close enough to $a$. - **(III)** has a fixed numerator $f(y) - Lg(y)$ and a denominator $g(x) \to \pm\infty$, so it tends to $0$. [/guided] [/step] [step:Bound the error by choosing $x$ sufficiently close to $a$] By the triangle inequality, \begin{align*} \left|\frac{f(x)}{g(x)} - L\right| \leq \underbrace{\left|\frac{f(y) - f(x)}{g(y) - g(x)} - L\right|}_{< \,\epsilon} \cdot \left|1 - \frac{g(y)}{g(x)}\right| + \frac{|f(y) - Lg(y)|}{|g(x)|}. \end{align*} Since $|g(x)| \to \infty$ as $x \to a^+$ while $g(y)$ and $f(y)$ are fixed, choose $\delta' \in (0, y - a)$ so that for all $x \in (a, a + \delta')$: - $\left|\frac{g(y)}{g(x)}\right| < 1$, which gives $\left|1 - \frac{g(y)}{g(x)}\right| \leq 1 + \left|\frac{g(y)}{g(x)}\right| < 2$, and - $\frac{|f(y) - Lg(y)|}{|g(x)|} < \epsilon$. Both conditions hold once $|g(x)| > \max\{|g(y)|, \, |f(y) - Lg(y)|/\epsilon\}$, which is achieved for $x$ close enough to $a$. For $x \in (a, a + \delta')$, the bound from the previous step on factor (I) combines with these estimates to give \begin{align*} \left|\frac{f(x)}{g(x)} - L\right| < \epsilon \cdot 2 + \epsilon = 3\epsilon. \end{align*} [guided] We now assemble the estimates. Applying the triangle inequality to the decomposition from the previous step: \begin{align*} \left|\frac{f(x)}{g(x)} - L\right| \leq \left|\frac{f(y) - f(x)}{g(y) - g(x)} - L\right| \cdot \left|1 - \frac{g(y)}{g(x)}\right| + \frac{|f(y) - Lg(y)|}{|g(x)|}. \end{align*} We need to bound both terms on the right. The first factor in the first term is already controlled: the Cauchy estimate gives $|\text{(I)}| < \epsilon$. For the remaining quantities, observe that $y$ is fixed, so $f(y)$, $g(y)$, and $f(y) - Lg(y)$ are all fixed [real numbers](/page/Real%20Numbers). The only variable quantity is $g(x)$, which satisfies $|g(x)| \to \infty$ as $x \to a^+$. Choose $\delta' \in (0, y - a)$ such that for all $x \in (a, a + \delta')$: \begin{align*} |g(x)| > \max\left\{|g(y)|, \; \frac{|f(y) - Lg(y)|}{\epsilon}\right\}. \end{align*} Such $\delta'$ exists because $|g(x)| \to \infty$ as $x \to a^+$. This single condition ensures both: - $|g(y)/g(x)| < |g(y)|/|g(y)| = 1$ (since $|g(x)| > |g(y)|$), so by the triangle inequality $|1 - g(y)/g(x)| \leq 1 + |g(y)/g(x)| < 2$. - $|f(y) - Lg(y)|/|g(x)| < \epsilon$ (since $|g(x)| > |f(y) - Lg(y)|/\epsilon$). Note that if $f(y) - Lg(y) = 0$, the second condition is automatic and any $\delta'$ ensuring $|g(x)| > |g(y)|$ suffices. Substituting into the bound: \begin{align*} \left|\frac{f(x)}{g(x)} - L\right| < \epsilon \cdot 2 + \epsilon = 3\epsilon. \end{align*} [/guided] [/step] [step:Conclude that $f(x)/g(x) \to L$] Since $\epsilon > 0$ was arbitrary, given any $\epsilon_0 > 0$ we apply the argument above with $\epsilon := \epsilon_0/3$, obtaining $\delta' > 0$ such that for all $x \in (a, a + \delta')$, \begin{align*} \left|\frac{f(x)}{g(x)} - L\right| < 3 \cdot \frac{\epsilon_0}{3} = \epsilon_0. \end{align*} This is the definition of $\lim_{x \to a^+} f(x)/g(x) = L$. $\blacksquare$ [/step]