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

This page is the drill floor. The parent note built the rule and its proof; here we hit every kind of problem the rule can face, so no exam surprises you. If a symbol appears you have not met, the parent note or Indeterminate Forms defines it.

The scenario matrix

Every problem below is a cell in this grid. "Direct" means L'Hôpital applies with no setup; "Convert" means you must reshape it into a fraction first.

Cell	Form	First move (WHY)	Example
A	$\frac{0}{0}$ direct	already a fraction of two zeros	(1)
B	$\frac{0}{0}$ repeated	new limit is still $\frac{0}{0}$ , apply again	(2)
C	$\frac{\infty}{\infty}$ direct	growth race, compare speeds	(3)
D	$0 \cdot \infty$ convert	rewrite $f \cdot g = \frac{f}{1/ g}$	(4)
E	$\infty - \infty$ convert	common denominator kills the subtraction	(5)
F	$1^{\infty}$ (log)	take $ln$ , drop the exponent	(6)
G	$0^{0}$ (log)	take $ln$ , exponent becomes a product	(7)
H	trap: looping	rule cycles forever → use algebra	(8)
I	word problem	build the limit yourself, then solve	(9)
J	exam twist: one-sided sign	the sign of the approach decides $\pm \infty$	(10)

Notation reminder, earned before use:

$x \to a lim$ reads "the value $x$ 's expression heads toward as $x$ creeps to $a$ " — see Limits — Definition & Laws.
$f^{'} (x)$ is the derivative: the instantaneous rate of change, the slope of the tangent line — see Linear Approximation & Tangent Lines.
$x \to 0^{+}$ means $x$ approaches $0$ from the right (only positive values); $x \to 0^{-}$ from the left.

Example 1 — Cell A: pure $\frac{0}{0}$

Worked example

$x \to 0 lim \frac{tan x - x}{x ^{3}}$ Forecast: guess before reading — is it $0$ , a finite nonzero number, or $\infty$ ? (The subtraction on top hints the numerator is tiny, tinier than $x$ but maybe comparable to $x^{3}$ .)

Check the form. At $x = 0$ : $tan 0 - 0 = 0$ on top, $0^{3} = 0$ on bottom. So $\frac{0}{0}$ . ✔ Why this step? You may never apply L'Hôpital without confirming both parts hit $0$ (or both $\infty$ ).
Differentiate top and bottom separately. $\frac{d}{d x} (tan x - x) = sec^{2} x - 1$ , and $\frac{d}{d x} x^{3} = 3 x^{2}$ . New limit: $\frac{sec ^{2} x - 1}{3 x ^{2}}$ . Why this step? The parent's proof says the ratio equals the ratio of speeds; that's $\frac{f ^{'}}{g ^{'}}$ , not the quotient rule.
Still $\frac{0}{0}$ ? At $0$ : $sec^{2} 0 - 1 = 1 - 1 = 0$ , bottom $0$ . Yes → apply again. $\frac{d}{d x} (sec^{2} x - 1) = 2 sec^{2} x tan x$ , bottom $6 x$ : $\frac{2 sec ^{2} x tan x}{6 x}$ . Why? A fresh $\frac{0}{0}$ is a fresh licence to reuse the rule.
Still $\frac{0}{0}$ . Split: $\frac{2 sec ^{2} x tan x}{6 x} = \frac{sec ^{2} x}{3} \cdot \frac{tan x}{x}$ . We know $\frac{t a n x}{x} \to 1$ and $sec^{2} 0 = 1$ , so limit $= \frac{1}{3}$ . Why factor instead of a third L'Hôpital? Recognising a known limit is cleaner and avoids messy derivatives.

Verify: Taylor gives $tan x = x + \frac{x ^{3}}{3} + \dots$ , so $tan x - x \approx \frac{x ^{3}}{3}$ , and dividing by $x^{3}$ gives $\frac{1}{3}$ . ✔ (see Taylor Series)

Example 2 — Cell B: $\frac{0}{0}$ repeated

Example 3 — Cell C: $\frac{\infty}{\infty}$ growth race

Look at the figure: it plots the numerator and denominator both racing to $\infty$ , but at wildly different rates.

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

Example 4 — Cell D: $0 \cdot \infty$

Example 5 — Cell E: $\infty - \infty$

Example 6 — Cell F: $1^{\infty}$ via logarithm

Example 7 — Cell G: $0^{0}$ via logarithm

Example 8 — Cell H: the looping trap (DON'T use L'Hôpital)

Example 9 — Cell I: word problem (build the limit yourself)

Worked example Continuous compounding

A bank pays annual rate $r$ but compounds $n$ times a year, so $1 grows to $(1 + \frac{r}{n})^{n}$ after one year. What happens as compounding becomes continuous, $n \to \infty$ ? Forecast: infinite compounding — does your dollar explode to $\infty$ , or settle at a finite number?

Set up the limit. We want $n \to \infty lim (1 + \frac{r}{n})^{n}$ . Why? "Continuous" = the number of compounding steps $\to \infty$ ; that is the limit.
Form: base $\to 1$ , exponent $\to \infty$ : $1^{\infty}$ (Cell F). Take $ln$ : $ln y = n ln (1 + \frac{r}{n})$ , form $\infty \cdot 0$ .
Convert: $ln y = \frac{ln ( 1 + r / n )}{1/ n}$ , $\frac{0}{0}$ . L'Hôpital in $n$ : top $^{'} = \frac{- r / n ^{2}}{1 + r / n}$ , bottom $^{'} = - \frac{1}{n ^{2}}$ . Ratio $= \frac{r}{1 + r / n} \to r$ .
Undo log. $ln y \to r \Rightarrow y \to e^{r}$ . Why exponentiate? The dollar value is $y$ , not $ln y$ .

Verify: with $r = 1$ this is the parent's Example (5), giving $e^{1} = e \approx 2.71828$ . Units: dollars per dollar (dimensionless growth factor). ✔

Example 10 — Cell J: exam twist, one-sided sign

Worked example

$x \to 0^{\pm} lim \frac{sin x}{x ^{3}}$ — treat both sides Forecast: it's $\frac{0}{0}$ , but the answer might not be a single number. Why might the two sides disagree?

Form: $sin 0 = 0$ , $0^{3} = 0$ : $\frac{0}{0}$ ✔.
L'Hôpital: $\frac{cos x}{3 x ^{2}}$ . Now top $\to cos 0 = 1 \neq = 0$ , bottom $\to 0^{+}$ (since $x^{2} > 0$ for either sign of $x$ ). Why note the bottom's sign? $3 x^{2}$ is always positive, so the fraction is $\frac{+ 1}{+ tiny} = + \infty$ from both sides.
Both-sided conclusion: $x \to 0 lim \frac{sin x}{x ^{3}} = + \infty$ . Why the twist matters: had the denominator been $x^{3}$ without the L'Hôpital step, its sign would flip ( $x^{3} < 0$ left, $> 0$ right) — but after one differentiation the denominator became $x^{2}$ , which is sign-blind. Always recheck the sign after each step.
Contrast case. For $x \to 0 lim \frac{sin x}{x ^{2}}$ : L'Hôpital gives $\frac{cos x}{2 x}$ , and $2 x$ does flip sign: $\to + \infty$ from the right, $\to - \infty$ from the left → the two-sided limit does not exist.

Verify: at $x = \pm 0.01$ : $sin (0.01) / (0.01)^{3} \approx 0.01/1 0^{- 6} = 1 0^{4} > 0$ both sides. ✔ And $sin (0.01) / (0.01)^{2} \approx + 100$ , while $sin (- 0.01) / (- 0.01)^{2} \approx - 100$ — opposite signs confirm no two-sided limit. ✔

Active recall

Recall Cover the answers

Before L'Hôpital, what must you always do? ::: Confirm the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
$lim_{x \to \infty} x^{2} / e^{x} = ?$ ::: $0$ (exponential beats any polynomial).
$lim_{x \to 0^{+}} x^{x} = ?$ ::: $1$ .
$lim_{x \to 0^{+}} (cos x)^{1/ x^{2}} = ?$ ::: $e^{- 1/2}$ .
When the rule loops (Example 8), what do you do? ::: Abandon it and use algebra.
Why can $lim_{x \to 0} sin x / x^{2}$ fail to exist while $sin x / x^{3}$ gives $+ \infty$ ? ::: After L'Hôpital the denominators are $2 x$ (sign flips) vs $3 x^{2}$ (always positive).

Connections

L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms (index 4.1.26) — parent: the rule and its proof.
Indeterminate Forms — the seven shapes of trouble enumerated in the matrix.
Taylor Series — every "Verify" via leading terms lives here.
Exponential & Logarithm Growth Rates — Examples 3, 4, 6, 7 are growth races.
Limits — Definition & Laws — one-sided limits and $\pm \infty$ (Example 10).
Linear Approximation & Tangent Lines — why $f^{'} / g^{'}$ is the right ratio.