L'Hôpital's Rule

Put $f(x) = \sin x$ and $g(x) = x$: then $f(0) = g(0) = 0$ and $g'(x) = 1 \neq 0$ — a $0/0$ form.

Both functions are continuous on $[0, x]$ and differentiable on $(0, x)$, so the hypotheses of Cauchy's mean value theorem are satisfied.

For $x > 0$, Cauchy's theorem on $[0, x]$ gives $\frac{\sin x - \sin 0}{x - 0} = \frac{\cos \xi_x}{1}$ for some $\xi_x \in (0, x)$.

Since $0 < \xi_x < x$, squeezing gives $\xi_x \to 0$ as $x \to 0^+$.

Cosine is continuous (differentiable implies continuous), so $\cos \xi_x \to \cos 0 = 1$.

That is L'Hôpital's rule for the $0/0$ case: the limit of the differentiated ratio equals the limit of the original ratio.

For $x < 0$, the same argument on $[x, 0]$ yields $\xi_x \in (x, 0)$ and likewise $\cos \xi_x \to 1$. Hence $\lim_{x \to 0}\frac{\sin x}{x} = 1$.

First the $\infty/\infty$ lemma: suppose $f, g$ are continuous on $[A_0, \infty)$ and differentiable on $(A_0, \infty)$ with $g' \neq 0$, $g(x) \to \infty$, and $\frac{f'(x)}{g'(x)} \to L$ (finite). Here $A_0$ is the fixed threshold from the derivative-limit hypothesis.

For a working $A \ge A_0$ and $x > A$, Cauchy's theorem on $[A, x]$ (continuity on $[A, x]$, differentiability on $(A, x)$) gives $\frac{f(x) - f(A)}{g(x) - g(A)} = \frac{f'(\xi_x)}{g'(\xi_x)}$ for some $\xi_x \in (A, x)$. Moreover $g' \neq 0$ forces $g$ to be strictly monotone by Rolle's theorem: if $g(x) = g(A)$ then Rolle would give a zero of $g'$.

Hence $g(x) \neq g(A)$ for $x > A$, justifying the division.

The key identity, shifted by $L$: $\frac{f(x)}{g(x)} - L = \left(\frac{f'(\xi_x)}{g'(\xi_x)} - L\right)\left(1 - \frac{g(A)}{g(x)}\right) + \frac{f(A) - L g(A)}{g(x)}$. Given $\varepsilon > 0$, first fix $A_0$ from the derivative-limit hypothesis so that $\left|\frac{f'(t)}{g'(t)} - L\right| < \frac{\varepsilon}{4}$ for $t > A_0$. Then choose the working $A \ge A_0$ (the role of $A$ is distinct from $A_0$: $A$ is the point frozen in the bounds below); then $\xi_x > A \ge A_0$ makes the first factor below $\frac{\varepsilon}{4}$.

Since $g(x) \to \infty$, for large $x$ we have $\left|\frac{g(A)}{g(x)}\right| < 1$, hence $\left|1 - \frac{g(A)}{g(x)}\right| \le 1 + \left|\frac{g(A)}{g(x)}\right| < 2$. The first term is at most $\frac{\varepsilon}{4} \cdot 2 = \frac{\varepsilon}{2}$, and the second term $\frac{|f(A) - L g(A)|}{|g(x)|}$ drops below $\frac{\varepsilon}{2}$ for large $x$.

Together, $\left|\frac{f(x)}{g(x)} - L\right| < \varepsilon$; with $\varepsilon$ arbitrary, $\frac{f(x)}{g(x)} \to L$ — L'Hôpital's rule for the $\infty/\infty$ case.

Apply this to $f(x) = \ln x$ and $g(x) = x$: $g'(x) = 1 \neq 0$, $g(x) \to \infty$, and $\frac{f'(x)}{g'(x)} = \frac{1/x}{1} = \frac{1}{x} \to 0$. Hence $\lim_{x \to \infty}\frac{\ln x}{x} = 0$.

For $x > 0$, $x \ln x = \frac{\ln x}{1/x}$. Put $f(x) = \ln x$ and $g(x) = 1/x$: then $f(x) \to -\infty$ and $g(x) \to +\infty$, an $\infty/\infty$ form, and $g'(x) = -\frac{1}{x^2} \neq 0$.

$\frac{f'(x)}{g'(x)} = \frac{1/x}{-1/x^2} = -x \xrightarrow[x \to 0^+]{} 0$. Restate problem 2's lemma in the flipped direction and re-verify its bounds: for fixed $A$ and $0 < x < A$, Cauchy's theorem on $[x, A]$ (again $g' \neq 0$ gives $g(x) \neq g(A)$ by Rolle) yields the same identity $\frac{f(x)}{g(x)} - L = \left(\frac{f'(\xi)}{g'(\xi)} - L\right)\left(1 - \frac{g(A)}{g(x)}\right) + \frac{f(A) - L g(A)}{g(x)}$ with $L = 0$ and $\xi \in (x, A)$.

As $x \to 0^+$, $|g(x)| = 1/x \to \infty$, so for the fixed $A$, $\left|1 - \frac{g(A)}{g(x)}\right| < 2$ for small $x$, and $\frac{|f(A) - L g(A)|}{|g(x)|} < \frac{\varepsilon}{2}$ for small $x$ — the fixed-$A$ terms are small because $|g(x)| \to \infty$. Choosing $A$ small first so $\left|\frac{f'(t)}{g'(t)} - L\right| < \frac{\varepsilon}{4}$ for $t \in (0, A)$ makes the first term at most $\frac{\varepsilon}{2}$, giving $\left|\frac{f(x)}{g(x)} - L\right| < \varepsilon$.

Hence $\lim_{x \to 0^+}\frac{\ln x}{1/x} = 0$.

Hence $\lim_{x \to 0^+} x \ln x = 0$. The numbers agree: at $x = 0.01$, $x\ln x = 0.01 \cdot \ln 0.01 \approx -0.0461$ and the derivative ratio is $-x = -0.01$, both tending to 0 as $x \to 0^+$.