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

WHAT is the problem?

When you plug $x \to a$ into $\dfrac{f(x)}{g(x)}$ and get $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ , the expression is indeterminate — the value is not determined by the form alone. Compare:

$\lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0}\frac{\sin x}{x^2}=\infty,\qquad \lim_{x\to 0}\frac{x^2}{\sin x}=0.$

All three look like $\frac{0}{0}$ yet give totally different answers. So $\frac{0}{0}$ is a question, not an answer.

WHY it's true — derivation by linear approximation

This is the "Feynman / derivation-from-scratch" view. Take the $\frac{0}{0}$ case with $f(a)=g(a)=0$ (define them so by continuity).

Step 1 — Linearise near $a$ . Near a point, a differentiable function looks like its tangent line: $f(x)\approx f(a)+f'(a)(x-a),\qquad g(x)\approx g(a)+g'(a)(x-a).$ Why this step? The whole power of derivatives is that a curve is its tangent line to first order — error vanishes faster than $(x-a)$ .

Step 2 — Use $f(a)=g(a)=0$ . The constant terms die: $\frac{f(x)}{g(x)}\approx \frac{f'(a)(x-a)}{g'(a)(x-a)}=\frac{f'(a)}{g'(a)}.$ Why this step? The $(x-a)$ factors are the shared "smallness". They cancel, leaving the ratio of speeds.

Step 3 — Take the limit. As $x\to a$ the approximation becomes exact: $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f'(a)}{g'(a)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}.$

Rigorous version (Cauchy Mean Value Theorem)

The linear-approximation argument secretly assumes $f',g'$ continuous. The airtight proof uses the Cauchy MVT: for $f,g$ on $[a,x]$ there is $c\in(a,x)$ with $\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(c)}{g'(c)}.$ With $f(a)=g(a)=0$ this is $\dfrac{f(x)}{g(x)}=\dfrac{f'(c)}{g'(c)}$ . As $x\to a$ , $c$ is squeezed to $a$ , so the limits match. Why this is stronger: no continuity of derivatives needed.

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

HOW to use it — and the indeterminate-form zoo

There are 7 indeterminate forms. Only $\frac{0}{0}$ and $\frac{\infty}{\infty}$ feed L'Hôpital directly; the rest must be converted.

Form	Trick to convert	Result type
$0/0$	use directly	—
$\infty/\infty$	use directly	—
$0\cdot\infty$	write $f\cdot g=\dfrac{f}{1/g}$	$0/0$ or $\infty/\infty$
$\infty-\infty$	common denominator / factor	$0/0$
$0^0,\ \infty^0,\ 1^\infty$	take $\ln$ : $y=f^g\Rightarrow \ln y=g\ln f$	$0\cdot\infty$

Worked examples

Common mistakes (Steel-man them)

Active recall

Recall Quick self-test (cover answers)

What two forms can you apply it to directly? ::: $\frac{0}{0}$ and $\frac{\infty}{\infty}$ .
Do you use the quotient rule? ::: No — differentiate numerator and denominator separately.
How do you handle $0^0$ ? ::: Take $\ln$ , turn into $0\cdot\infty$ , then $0/0$ .
What must be true for the conclusion to hold? ::: The limit of $f'/g'$ must exist (or be $\pm\infty$ ).

Recall Feynman: explain to a 12-year-old

Two snails start at the same line and we ask "after a tiny moment, who is further?" Since they both started at zero, the only thing that decides it is how fast each crawls — that's the slope, the derivative. So instead of comparing their tiny distances (which are both basically zero — confusing!), we just compare their speeds. That swap, distances → speeds, is L'Hôpital's rule.

Flashcards

State L'Hôpital's rule for

0/0

f(a)=g(a)=0

g'\neq0

near

a

, and

\lim f'/g'

exists, then

\lim_{x\to a} f/g = \lim_{x\to a} f'/g'

Why does linear approximation explain the rule?

Near

a

f\approx f'(a)(x-a)

and

g\approx g'(a)(x-a)

; the

(x-a)

cancel leaving

f'(a)/g'(a)

Which theorem makes the proof rigorous?

The Cauchy Mean Value Theorem.

List the 7 indeterminate forms.

0/0,\ \infty/\infty,\ 0\cdot\infty,\ \infty-\infty,\ 0^0,\ \infty^0,\ 1^\infty

How do you convert

0\cdot\infty

Write

f\cdot g = f/(1/g)

to make

0/0

\infty/\infty

How do you handle

1^\infty,0^0,\infty^0

Take natural log to get

0\cdot\infty

, convert to a fraction, apply the rule, then exponentiate.

Common error: applying which rule by mistake?

The quotient rule — instead differentiate top and bottom independently.

\lim_{x\to0}(e^x-1)/x = ?

1

\lim_{x\to\infty}(\ln x)/x=?

0

\lim_{x\to\infty}(1+1/x)^x=?

e

\lim f'/g'

does not exist, what can you conclude about

\lim f/g

Nothing — the rule is one-directional;

\lim f/g

may still exist.

Connections

Mean Value Theorem — Cauchy MVT is the engine of the proof.
Linear Approximation & Tangent Lines — the intuitive derivation.
Taylor Series — generalises: ratio of leading nonzero Taylor terms gives the limit.
Indeterminate Forms — the broader classification.
Limits — Definition & Laws — what a limit must satisfy.
Exponential & Logarithm Growth Rates — example (3) compares growth speeds.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab limit lete waqt $\frac{0}{0}$ ya $\frac{\infty}{\infty}$ aata hai, toh yeh koi answer nahi hai — yeh ek sawaal hai. Top aur bottom dono zero ki taraf ja rahe hain, lekin asli sawaal yeh hai ki kaun kitni speed se zero ki taraf ja raha hai. Aur speed ka matlab hai derivative. Isliye L'Hôpital bolta hai: limit of $f/g$ ko badal do limit of $f'/g'$ se. Bas ek dhyan rakho — yeh quotient rule nahi hai; upar aur neeche ko alag-alag differentiate karna hai.

Iska proof simple intuition se aata hai: koi bhi smooth function apne point ke paas apni tangent line jaisa dikhta hai, yani $f(x)\approx f'(a)(x-a)$ aur $g(x)\approx g'(a)(x-a)$ (kyunki $f(a)=g(a)=0$ ). Ab dono mein $(x-a)$ cancel ho jata hai aur bachta hai $f'(a)/g'(a)$ . Bilkul jaise do snail same line se start karein — thodi der baad jo zyada tej crawl karega wahi aage hoga, aur "tej" ka matlab slope yani derivative.

Baaki forms — $0\cdot\infty$ , $\infty-\infty$ , $0^0$ , $1^\infty$ , $\infty^0$ — pehle inhe $0/0$ ya $\infty/\infty$ mein convert karna padta hai. $0\cdot\infty$ ko fraction bana do ( $f/(1/g)$ ); aur power waale forms ( $1^\infty$ type) mein pehle log lo, jisse exponent neeche aa jaata hai. Phir hi rule lagao.

Ek important warning: hamesha pehle form check karo, aur agar $f'/g'$ ka limit exist nahi karta toh tum yeh nahi keh sakte ki original limit bhi nahi hai. Aur kabhi-kabhi rule loop maarta rehta hai (jaise $e^x/\sqrt{e^{2x}+1}$ ) — tab algebra use karo, zid mat karo!

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

WHAT is the problem?

WHY it's true — derivation by linear approximation

Rigorous version (Cauchy Mean Value Theorem)

HOW to use it — and the indeterminate-form zoo

Worked examples

Common mistakes (Steel-man them)

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections