L'Hôpital's Rule Preview

Learning Objectives

By the end of this section, you will be able to:

1. Recognise the seven indeterminate forms and explain why they defy direct substitution.
2. State L'Hôpital's Rule and describe the hypotheses under which it is legal to apply.
3. Interpret the rule geometrically — two curves kissing the axis with slopes whose ratio is the limit.
4. Compute limits of the forms 0/0 and ∞/∞ using the rule, and convert 0 · ∞, ∞ − ∞, 0⁰, 1^∞, ∞⁰ into one of those two forms.
5. Verify the rule numerically in Python and automate it with PyTorch's autograd.

The Problem: When Algebra Runs Out

"What is 0 divided by 0?" The honest answer is: it depends on how both parts got to zero.

Up to this point, computing a limit at a point $x = c$ has been a two-step choreography: plug in $c$, and if the result is a finite number, that is the limit. The trouble is that a quotient like

hands back $0/0$ the moment you substitute. The numerator and denominator both vanish, and the ratio is whatever you want it to be — $0/0$ carries no information. Yet the limit clearly exists, because if you plot the curve near $x = 0$ you see it settle comfortably at $3/5$.

In 1696, Guillaume de L'Hôpital published the first calculus textbook, and inside it was a rule (learned from his tutor, Johann Bernoulli) that lets us substitute the derivatives of the numerator and denominator when the originals both go to zero. The rule is astonishingly cheap to state, astonishingly powerful in practice, and leans on every piece of machinery we have built so far: limits, continuity, and the derivative's meaning as a local rate.

What Indeterminate Forms Actually Mean

Not every suspicious-looking expression is indeterminate. A limit that produces$\;0/5$ is simply $0$. A limit that produces $5/0$ with the denominator approaching 0 from the positive side is $+\infty$. Indeterminate forms are the ones where different underlying rates produce different answers.

Intuition: Two Cars Racing to Zero

Think of $f(x)$ and $g(x)$ as positions of two cars on the number line. At time $x = c$ they both reach the finish line $y = 0$. The ratio $f(x)/g(x)$ is the distance ratio between the two cars just before they arrive.

This is exactly the content of the rule: the indeterminate ratio of positions is determined by the ratio of the rates at which they reach zero. The derivative $f'(c)$ literally means "how fast f is changing at c" — which for a function vanishing at c is the same as "how fast f is approaching zero at c".

The Rule, Stated Precisely

Why It Works — A Linear Approximation Proof

The clearest way to see the rule is to Taylor-expand. Because $f$ and $g$ are differentiable at $c$ with $f(c) = g(c) = 0$, for $x$ near $c$:

where the remainders $R_f, R_g$ are $o(x - c)$ — they vanish faster than the linear term. Dividing,

Because $R_f(x)/(x-c) \to 0$ and $R_g(x)/(x-c) \to 0$ as $x \to c$, the middle expression tends to $f'(c)/g'(c)$. That is the rule. A more careful proof uses the Cauchy Mean Value Theorem — which is the main-course version of the idea we'll cover in a later chapter — but the linear-approximation picture is the honest intuition.

Interactive: Compare f/g and f'/g'

Below you can step through six indeterminate-form limits side by side. The red plot is the naive ratio $f(x)/g(x)$; the green plot is the ratio of derivatives $f'(x)/g'(x)$. Drag the zoom slider to shrink the window around the trouble point. Watch how the red curve has to crawl toward the answer while the green curve sits on it from the start.

Try this experimental loop:

1. Start with sin(x)/x at x → 0. Zoom all the way in — the red curve wobbles numerically near 0, but the green cos(x)/1 is perfectly flat at 1.
2. Switch to (1 − cos x)/x². Zoom in. The red crawl is slow because the numerator is quadratic; but $\sin(x)/(2x)$ is still 0/0, so the green curve itself is an indeterminate ratio until you apply the rule a second time (mentally) to land on 1/2.
3. Jump to x/eˣ at x → ∞. Now you zoom on a very large window. The red ratio falls from 1/eˣ quickly; the green curve $1/e^x$ is already the derivative answer.
4. Pick x · ln(x) at x → 0⁺. This is a 0·∞ form. The rewrite$\;x/(1/\ln x)$ converts it to 0/0 and shows both ratios climbing to 0.

Worked Example: lim sin(3x)/sin(5x)

We will compute $\displaystyle \lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$ three ways: as a direct substitution (fails), via L'Hôpital, and via a small Taylor argument. Do each step on paper — it builds the reflex for recognising when the rule applies and when it does not.

Handling ∞/∞ and Other Forms

∞ / ∞ directly: x / eˣ as x → ∞

Both $x$ and $e^x$ blow up, so direct evaluation gives $\infty/\infty$. Differentiate:

One L'Hôpital call collapsed a hard question into an easy one. This also proves the classical fact that $e^x$ dominates every polynomial — apply the rule $n$ times to $x^n/e^x$ and you end up with $n!/e^x \to 0$.

0 · ∞: x ln(x) as x → 0⁺

A product of 0 and $-\infty$ is ambiguous. Rewrite it as a quotient to force a 0/0 or ∞/∞ form, whichever is easier to differentiate:

Apply L'Hôpital:

Notice the freedom in the rewrite: we could also have used $\tfrac{x}{1/\ln x}$ (a 0/0 form), but the resulting derivative is uglier. Part of the skill is choosing the rewrite that leads to the cleanest derivative.

1^∞ and ∞⁰: take the logarithm first

For powers whose base and exponent both misbehave, take $\ln$ to convert a product of a log and another ambiguous factor, then use the techniques above. The classical example is

Let $y = (1 + 1/n)^n$ so $\ln y = n \ln(1 + 1/n)$, an $\infty \cdot 0$ form. Rewrite:

The resulting limit is $1$, so $\ln y \to 1$ and $y \to e^1 = e$. The same trick handles 0⁰, 1^∞, and ∞⁰ uniformly.

Python: Verifying L'Hôpital Numerically

The rule is a theorem, but a computer can illustrate it beautifully. Below we compute $\sin(3x)/\sin(5x)$ at shrinking $x$ values and watch it converge to the L'Hôpital answer $3/5$. We then compute the answer directly by plugging $x = 0$ into $f'/g'$ — zero shrinking required.

1import math
2
3def f(x):
4    """Numerator — vanishes at x = 0."""
5    return math.sin(3 * x)
6
7def g(x):
8    """Denominator — also vanishes at x = 0."""
9    return math.sin(5 * x)
10
11def f_prime(x):
12    """Analytic derivative of sin(3x) is 3 cos(3x)."""
13    return 3 * math.cos(3 * x)
14
15def g_prime(x):
16    """Analytic derivative of sin(5x) is 5 cos(5x)."""
17    return 5 * math.cos(5 * x)
18
19# --- 1) Naive evaluation: shrink x toward 0 and watch f/g. -----------------
20print("Naive limit of f(x)/g(x) as x -> 0")
21print(f"{'x':>10}  {'f(x)':>12}  {'g(x)':>12}  {'f(x)/g(x)':>14}")
22print("-" * 54)
23for x in [0.5, 0.1, 0.01, 0.001, 1e-4]:
24    fx = f(x)
25    gx = g(x)
26    print(f"{x:>10.4g}  {fx:>12.6f}  {gx:>12.6f}  {fx / gx:>14.8f}")
27
28# --- 2) L'Hopital shortcut: evaluate f'/g' at x = 0 directly. --------------
29print()
30print("L'Hopital: evaluate ratio of derivatives AT x = 0")
31limit = f_prime(0) / g_prime(0)        # 3*cos(0) / 5*cos(0) = 3/5
32print(f"f'(0) / g'(0) = {f_prime(0)}/{g_prime(0)} = {limit:.6f}")
33
34# --- 3) Sanity check: the two approaches agree in the limit. ---------------
35print()
36print("Absolute error between naive ratio and 3/5")
37for x in [0.1, 0.01, 0.001, 1e-4]:
38    err = abs(f(x) / g(x) - limit)
39    print(f"x = {x:>7.0e}   |f/g - 3/5| = {err:.3e}")

PyTorch: Using Autograd as an L'Hôpital Engine

Hand-differentiating $\sin(3x)$ is easy. Hand-differentiating a 100-term loss function in a neural net is not. Autograd does it for us, which means we can build a fully automatic L'Hôpital evaluator: declare $f$ and $g$, call .backward() twice, divide, done.

1import torch
2
3def f(x):
4    return torch.sin(3 * x)
5
6def g(x):
7    return torch.sin(5 * x)
8
9# 1) A point where BOTH functions vanish. We probe "just off zero"
10#    because autograd at a leaf tensor only needs a place to evaluate;
11#    but evaluating exactly at 0 works too, since f and g are analytic.
12x = torch.tensor(0.0, requires_grad=True)
13
14# 2) Evaluate f and g once. This builds two autograd graphs that
15#    share x as the leaf.
16y_f = f(x)
17y_g = g(x)
18
19# 3) Backprop through f separately from g. We zero out x.grad in between
20#    because PyTorch accumulates gradients by default.
21y_f.backward(retain_graph=True)
22f_prime = x.grad.item()
23x.grad = None                      # clear accumulator
24
25y_g.backward()
26g_prime = x.grad.item()
27
28print(f"f(0)  = {y_f.item():.6f}   (should be 0)")
29print(f"g(0)  = {y_g.item():.6f}   (should be 0)")
30print(f"f'(0) = {f_prime:.6f}")
31print(f"g'(0) = {g_prime:.6f}")
32print(f"L'Hopital limit = f'(0) / g'(0) = {f_prime / g_prime:.6f}")
33
34# 4) Sanity check: sweep x and compare f(x)/g(x) with the L'Hopital answer.
35print()
36print("Autograd L'Hopital vs numerical ratio")
37with torch.no_grad():
38    for xv in [0.1, 0.01, 0.001, 1e-4]:
39        xt = torch.tensor(xv)
40        naive = (f(xt) / g(xt)).item()
41        err = abs(naive - f_prime / g_prime)
42        print(f"x = {xv:>7.0e}  naive = {naive:.8f}  err = {err:.2e}")

Where L'Hôpital Unlocks Real Problems

Common Pitfalls

Summary

L'Hôpital's Rule is a bridge between two pillars of calculus: the limit and the derivative. When direct substitution produces an indeterminate form, the rule says: the answer is controlled by the rate at which numerator and denominator approach their limiting values. Formally,

whenever the hypotheses (indeterminate form, differentiability, $g' \neq 0$, and existence of the right-hand limit) hold.

Key Takeaways

1. L'Hôpital is a statement about rates: the ratio of values at the trouble point equals the ratio of their derivatives there.
2. The rule applies only to 0/0 and ∞/∞ forms directly; every other indeterminate form must be rewritten into one of these first.
3. If the new ratio is still indeterminate, apply the rule again — or switch to a Taylor expansion for clarity.
4. The proof is just a linear approximation: $f(x) \approx f'(c)(x-c)$ and $g(x) \approx g'(c)(x-c)$, then cancel $(x-c)$.
5. In practice, PyTorch's autograd turns L'Hôpital into a two-line procedure — which is effectively what every deep-learning optimiser does, billions of times per second, at scale.

Coming Next: Chapter 3 turns the ε–δ language into a property of functions themselves: continuity. We will see why continuous functions are the sandbox on which every theorem about derivatives and integrals is built — and how the smallest break in a curve unravels entire theorems.