L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms: Visual walkthrough

This is a companion to the main note. Here the pictures carry the argument.

Step 1 — What does $\frac{0}{0}$ even look like?

WHAT. We have two functions. Call the top one $f$ and the bottom one $g$ . We want the value of $\frac{f ( x )}{g ( x )}$ as $x$ slides toward a special number $a$ . The symbol $x \to a$ just means " $x$ creeps closer and closer to $a$ ". The trouble: at $x = a$ both functions equal zero.

WHY it's a problem. Zero divided by zero is not a number — it is a question. Look at the figure: two different curves can both pass through zero at the same spot, yet their ratio near that spot can be anything. The form alone tells you nothing.

PICTURE. Two curves, $f$ (blue) and $g$ (orange), both crossing the horizontal axis at the same point $x = a$ . The red dot marks the shared zero.

Figure — L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms

Step 2 — Zoom in: a curve is its tangent line

WHAT. Pick either curve and zoom your camera in hard on the point $x = a$ . As you zoom, the curve stops looking curvy and looks like a straight line — its tangent line. This is the idea of Linear Approximation & Tangent Lines.

WHY this tool and not another? We need a simple replacement for a complicated curve. The tangent line is the best straight-line copy of the curve at that point — it matches the height and the slope. And a straight line through a known point is trivial to write down. This is exactly the tool that turns "hard curve ratio" into "easy line ratio".

PICTURE. The blue curve on the left looks bent. On the right, a zoomed box shows the same curve almost perfectly overlapping its tangent line (dashed). The green segment shows the slope.

The same copy works for $g$ : $g (x) \approx g (a) + g^{'} (a) (x - a) .$

Step 3 — Kill the constant terms

WHAT. Substitute our two tangent-line copies into the ratio, then use the fact from Step 1 that $f (a) = 0$ and $g (a) = 0$ .

WHY. The constant terms $f (a)$ and $g (a)$ are the heights at $a$ . Both are zero here — that is the whole meaning of " $\frac{0}{0}$ ". So they simply disappear, and what remains is pure slope times step.

PICTURE. Both tangent lines now pass through the origin-height (the axis). Their whole story is now just their slopes — the blue line steeper than the orange, say.

$\frac{f ( x )}{g ( x )} \approx \frac{f ( a ) = 0 + f ^{'} ( a ) ( x - a )}{= 0 g ( a ) + g ^{'} ( a ) ( x - a )} = \frac{f ^{'} ( a ) ( x - a )}{g ^{'} ( a ) ( x - a )}$

The struck-out $f (a), g (a)$ ::: gone, because both are zero at $a$ .
What is left ::: slope times the same small step, on top and bottom.

Step 4 — Cancel the shared smallness

WHAT. The factor $(x - a)$ appears on top and bottom. It is the same little number in both places, so it cancels.

WHY this is the punchline. The $(x - a)$ is the shared shrinking — it is the "both racing to zero" part. Once you notice it is identical on top and bottom, it cannot decide the answer. Strip it away and the only thing left standing is the ratio of slopes — the ratio of speeds.

PICTURE. The $(x - a)$ on top and bottom lit up in the same color, with a slash through both — leaving just $f^{'} (a) / g^{'} (a)$ .

$\frac{f ^{'} ( a ) ( x - a )}{g ^{'} ( a ) ( x - a )} = \frac{f ^{'} ( a )}{g ^{'} ( a )}$

Step 5 — Take the limit for real

WHAT. Everything above used " $\approx$ ". Now let $x$ actually slide into $a$ . As it does, the tangent-line copies become exact (their gap vanished faster than $(x - a)$ ), so the approximation becomes an equality.

WHY. We wanted the limit, not an approximation. The limit is where the tangent-line copy and the true curve agree perfectly.

PICTURE. A sequence of ever-closer $x$ values marching toward $a$ ; the ratio's value settling onto the horizontal dashed line at height $f^{'} (a) / g^{'} (a)$ .

Step 6 — The honest fix: Cauchy's Mean Value Theorem

WHAT. Steps 2–5 quietly assumed the slopes $f^{'}, g^{'}$ are themselves smooth (continuous). To be airtight we replace "slope at $a$ " with "slope somewhere between $a$ and $x$ ", using the Cauchy Mean Value Theorem.

WHY. The Cauchy MVT hands us an exact equation — no " $\approx$ " anywhere — that says: somewhere between $a$ and $x$ there is a point $c$ where the ratio of the total changes equals the ratio of the instantaneous slopes.

PICTURE. The interval from $a$ to $x$ on the axis, with a point $c$ (green) trapped strictly inside. An arrow shows $c$ getting squeezed toward $a$ as $x \to a$ .

$\frac{f ( x ) - f ( a ) 0}{g ( x ) - 0 g ( a )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}, c between a and x$

Since $f (a) = g (a) = 0$ the left side is exactly $\frac{f ( x )}{g ( x )}$ . As $x \to a$ , the trapped point $c$ has nowhere to go but $a$ , so $\frac{f ^{'} ( c )}{g ^{'} ( c )} \to lim \frac{f ^{'}}{g ^{'}}$ . Same conclusion — no assumption that the derivatives are continuous.

Step 7 — Edge case: what if the rule loops forever?

WHAT. The rule only earns its answer if the right-hand limit $lim f^{'} / g^{'}$ actually exists (finite or $\pm \infty$ ). Sometimes differentiating just gives you a new fraction as bad as the old one, forever.

WHY show this. A reader who trusts the rule blindly will spin in circles. Consider $lim_{x \to \infty} \frac{e ^{x}}{e ^{2 x} + 1} .$ Each application flips the fraction over and never simplifies. The escape is algebra: divide top and bottom by $e^{x}$ .

PICTURE. A loop arrow (the rule cycling back to itself) beside a straight arrow (algebra) that lands on the answer $1$ .

$\frac{e ^{x}}{e ^{2 x} + 1} = \frac{1}{1 + e ^{- 2 x}} x \to \infty \frac{1}{1 + 0} = 1$

$e^{- 2 x}$ ::: shrinks to $0$ as $x \to \infty$ (see Exponential & Logarithm Growth Rates).
The whole thing settles cleanly on $1$ — no looping.

Step 8 — Edge case: forms that must be converted first

WHAT. Only $\frac{0}{0}$ and $\frac{\infty}{\infty}$ feed the rule directly. The other indeterminate forms are disguises — reshape them into a fraction first.

WHY. L'Hôpital differentiates a ratio. If your problem is a product $0 \cdot \infty$ or a power $1^{\infty}$ , there is no ratio yet — you must build one.

PICTURE. A small flow map: each stray form arrow-ing into $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

The one-picture summary

This single figure compresses the whole journey: two curves through a shared zero (Step 1) → each replaced by its tangent line (Step 2) → constants die (Step 3) → the shared $(x - a)$ cancels (Step 4) → limit becomes the slope ratio (Step 5).

Recall Feynman retelling — say it plain

Two curves both dive through zero at the same spot, so their ratio is the maddening $\frac{0}{0}$ . Instead of staring at two vanishing heights, I zoom in until each curve becomes its own straight tangent line — a line I can write with just its slope and a step $(x - a)$ . Because both start at height zero, the constant terms vanish; both lines are just slope times the same little step. That little step $(x - a)$ sits on top and bottom identically, so it cancels — it was the "both racing to zero" part and it can't decide the winner. What survives is purely the ratio of slopes, $f^{'} (a) / g^{'} (a)$ : the ratio of speeds. Cauchy's theorem makes this exact by pinning the slopes at a point $c$ squeezed to $a$ . And I stay honest: the trick only pays off if that speed-ratio actually settles down; if it loops, I reach for algebra instead.

Connections

Linear Approximation & Tangent Lines — Steps 2–4 are literally this idea in action.
Mean Value Theorem — Step 6's Cauchy MVT makes the picture rigorous.
Taylor Series — the next level: compare leading nonzero terms, not just first slopes.
Indeterminate Forms — Step 8's zoo of disguises.
Exponential & Logarithm Growth Rates — powers Step 7's edge case.
Limits — Definition & Laws — what " $lim$ " and " $x \to a$ " actually mean.