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

4.1.26 · D3 · Maths › Calculus I — Limits & Derivatives › L'Hôpital's rule — proof using linear approximation, 0 - 0,

Yeh page drill floor hai. parent note mein rule aur uska proof banaya gaya tha; yahan hum har tarah ki problem solve karte hain jo rule ko face karni pad sakti hai, taaki koi bhi exam tumhe surprise na kar sake. Agar koi symbol naya lage, toh parent note ya Indeterminate Forms mein uski definition milegi.

The scenario matrix

Neeche ki har problem is grid ka ek cell hai. "Direct" ka matlab hai L'Hôpital bina kisi setup ke apply hota hai; "Convert" ka matlab hai pehle use fraction ki shape mein reshape karna hoga.

Cell	Form	Pehla kadam (KYU)	Example
A	$\frac{0}{0}$ direct	already do zeros ka fraction hai	(1)
B	$\frac{0}{0}$ repeated	naya limit phir bhi $\frac{0}{0}$ hai, dobara apply karo	(2)
C	$\frac{\infty}{\infty}$ direct	growth race, speeds compare karo	(3)
D	$0 \cdot \infty$ convert	$f \cdot g = \frac{f}{1/ g}$ likhkar convert karo	(4)
E	$\infty - \infty$ convert	common denominator subtraction khatam kar deta hai	(5)
F	$1^{\infty}$ (log)	$ln$ lo, exponent neeche aa jaata hai	(6)
G	$0^{0}$ (log)	$ln$ lo, exponent ek product ban jaata hai	(7)
H	trap: looping	rule hamesha cycle karta hai → algebra use karo	(8)
I	word problem	pehle limit khud banao, phir solve karo	(9)
J	exam twist: one-sided sign	approach ka sign decide karta hai $\pm \infty$	(10)

Notation reminder, pehle samjho phir use karo:

$x \to a lim$ padho "jab $x$ creep karta hai $a$ ki taraf, expression us value ki taraf jaata hai" — dekho Limits — Definition & Laws.
$f^{'} (x)$ derivative hai: change ki instantaneous rate, tangent line ki slope — dekho Linear Approximation & Tangent Lines.
$x \to 0^{+}$ matlab $x$ , $0$ ke paas right side se jaata hai (sirf positive values); $x \to 0^{-}$ matlab left side se.

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

Worked example

$x \to 0 lim \frac{tan x - x}{x ^{3}}$ Forecast: padhne se pehle guess karo — kya yeh $0$ hai, ek finite nonzero number, ya $\infty$ ? (Upar ki subtraction hint deti hai ki numerator tiny hai, $x$ se bhi tiny, lekin shayad $x^{3}$ se comparable.)

Form check karo. $x = 0$ par: $tan 0 - 0 = 0$ upar, $0^{3} = 0$ neeche. Toh $\frac{0}{0}$ . ✔ Yeh step kyun? L'Hôpital tab tak kabhi apply nahi karte jab tak confirm na ho ki dono parts $0$ hit karte hain (ya dono $\infty$ ).
Top aur bottom ko alag-alag differentiate karo. $\frac{d}{d x} (tan x - x) = sec^{2} x - 1$ , aur $\frac{d}{d x} x^{3} = 3 x^{2}$ . Naya limit: $\frac{sec ^{2} x - 1}{3 x ^{2}}$ . Yeh step kyun? Parent ka proof kehta hai ratio, speeds ke ratio ke barabar hai; woh hai $\frac{f ^{'}}{g ^{'}}$ , quotient rule nahi.
Abhi bhi $\frac{0}{0}$ ? $0$ par: $sec^{2} 0 - 1 = 1 - 1 = 0$ , bottom $0$ . Haan → dobara apply karo. $\frac{d}{d x} (sec^{2} x - 1) = 2 sec^{2} x tan x$ , bottom $6 x$ : $\frac{2 sec ^{2} x tan x}{6 x}$ . Kyun? Nayi $\frac{0}{0}$ matlab rule reuse karne ka naya license.
Phir bhi $\frac{0}{0}$ . Split karo: $\frac{2 sec ^{2} x tan x}{6 x} = \frac{sec ^{2} x}{3} \cdot \frac{tan x}{x}$ . Hum jaante hain $\frac{t a n x}{x} \to 1$ aur $sec^{2} 0 = 1$ , toh limit $= \frac{1}{3}$ . Teesra L'Hôpital lagane ki jagah factor kyun? Ek jaani-pahchani limit ko pehchanna zyada saaf hota hai aur messy derivatives se bachata hai.

Verify: Taylor deta hai $tan x = x + \frac{x ^{3}}{3} + \dots$ , toh $tan x - x \approx \frac{x ^{3}}{3}$ , aur $x^{3}$ se divide karne par $\frac{1}{3}$ milta hai. ✔ (dekho Taylor Series)

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

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

Figure dekho: isme numerator aur denominator dono $\infty$ ki taraf race kar rahe hain, lekin bilkul alag rates par.

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: $1^{\infty}$ logarithm ke zariye

Worked example

$x \to 0^{+} lim (cos x)^{1/ x^{2}}$ Forecast: base $\to 1$ , exponent $\to \infty$ . Ek $1^{\infty}$ — answer $1$ hai, $e$ hai, ya $1$ se kam kuch?

Form: $cos 0 = 1$ , exponent $1/ x^{2} \to \infty$ . $1^{\infty}$ — indeterminate kyunki " $1$ -ish ko bahut bade power par raise karna" kuch bhi ho sakta hai.
$ln$ lo. Maano $y = (cos x)^{1/ x^{2}}$ , tab $ln y = \frac{ln ( cos x )}{x ^{2}}$ . $ln$ kyun? Exponents differentiation ke liye tab tak invisible hote hain jab tak $ln$ unhe neeche product/quotient mein na kheeche — parent Example (5) dekho.
$ln y$ ki form: $ln (cos 0) = ln 1 = 0$ , $x^{2} = 0$ se divide: $\frac{0}{0}$ . L'Hôpital: top $^{'} = \frac{- sin x}{cos x} = - tan x$ , bottom $^{'} = 2 x$ → $\frac{- tan x}{2 x} \to - \frac{1}{2}$ ( $tan x / x \to 1$ use karke).
Log undo karo. $ln y \to - \frac{1}{2} \Rightarrow y \to e^{- 1/2} = \frac{1}{e}$ . Exponentiate kyun? Humne $ln y$ solve kiya tha; original limit $e^{(l n y)}$ hai.

Verify: $e^{- 1/2} \approx 0.6065$ . $x = 0.1$ par: $(cos 0.1)^{100} = (0.995004)^{100} \approx 0.6062$ . ✔

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

Example 8 — Cell H: looping trap (L'Hôpital MAT use karo)

Example 9 — Cell I: word problem (limit khud banao)

Worked example Continuous compounding

Ek bank annual rate $r$ deta hai lekin saal mein $n$ baar compound karta hai, toh $1 ek saal baad $(1 + \frac{r}{n})^{n}$ ban jaata hai. Kya hoga jab compounding continuous ho jaaye, $n \to \infty$ ? Forecast: infinite compounding — kya tumhara dollar $\infty$ tak explode hoga, ya kisi finite number par settle karega?

Limit set up karo. Hum chahte hain $n \to \infty lim (1 + \frac{r}{n})^{n}$ . Kyun? "Continuous" = compounding steps ki sankhya $\to \infty$ ; woh hi limit hai.
Form: base $\to 1$ , exponent $\to \infty$ : $1^{\infty}$ (Cell F). $ln$ lo: $ln y = n ln (1 + \frac{r}{n})$ , form $\infty \cdot 0$ .
Convert: $ln y = \frac{ln ( 1 + r / n )}{1/ n}$ , $\frac{0}{0}$ . $n$ mein L'Hôpital: top $^{'} = \frac{- r / n ^{2}}{1 + r / n}$ , bottom $^{'} = - \frac{1}{n ^{2}}$ . Ratio $= \frac{r}{1 + r / n} \to r$ .
Log undo karo. $ln y \to r \Rightarrow y \to e^{r}$ . Exponentiate kyun? Dollar ki value $y$ hai, $ln y$ nahi.

Verify: $r = 1$ ke saath yeh parent ka Example (5) hai, jo deta hai $e^{1} = e \approx 2.71828$ . Units: dollars per dollar (dimensionless growth factor). ✔

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

Worked example

$x \to 0^{\pm} lim \frac{sin x}{x ^{3}}$ — dono sides treat karo Forecast: yeh $\frac{0}{0}$ hai, lekin answer ek single number nahi ho sakta. Dono sides kyun disagree kar sakti hain?

Form: $sin 0 = 0$ , $0^{3} = 0$ : $\frac{0}{0}$ ✔.
L'Hôpital: $\frac{cos x}{3 x ^{2}}$ . Ab top $\to cos 0 = 1 \neq = 0$ , bottom $\to 0^{+}$ (kyunki $x^{2} > 0$ $x$ ke kisi bhi sign ke liye). Bottom ka sign kyun note karein? $3 x^{2}$ hamesha positive hai, toh fraction $\frac{+ 1}{+ tiny} = + \infty$ hai dono sides se.
Both-sided conclusion: $x \to 0 lim \frac{sin x}{x ^{3}} = + \infty$ . Twist kyun matter karta hai: agar denominator $x^{3}$ hota bina L'Hôpital step ke, toh uska sign flip ho jaata ( $x^{3} < 0$ left, $> 0$ right) — lekin ek differentiation ke baad denominator $x^{2}$ ban gaya, jo sign-blind hai. Har step ke baad sign hamesha recheck karo.
Contrast case. $x \to 0 lim \frac{sin x}{x ^{2}}$ ke liye: L'Hôpital deta hai $\frac{cos x}{2 x}$ , aur $2 x$ ka sign flip hota hai: $\to + \infty$ right se, $\to - \infty$ left se → two-sided limit exist nahi karta.

Verify: $x = \pm 0.01$ par: $sin (0.01) / (0.01)^{3} \approx 0.01/1 0^{- 6} = 1 0^{4} > 0$ dono sides. ✔ Aur $sin (0.01) / (0.01)^{2} \approx + 100$ , jabki $sin (- 0.01) / (- 0.01)^{2} \approx - 100$ — opposite signs confirm karte hain ki koi two-sided limit nahi hai. ✔

Active recall

Recall Answers cover karo

L'Hôpital se pehle tumhe hamesha kya karna chahiye? ::: Confirm karo ki form $\frac{0}{0}$ ya $\frac{\infty}{\infty}$ hai.
$lim_{x \to \infty} x^{2} / e^{x} = ?$ ::: $0$ (exponential kisi bhi polynomial ko beat karta hai).
$lim_{x \to 0^{+}} x^{x} = ?$ ::: $1$ .
$lim_{x \to 0^{+}} (cos x)^{1/ x^{2}} = ?$ ::: $e^{- 1/2}$ .
Jab rule loop kare (Example 8), toh kya karo? ::: Chhod do aur algebra use karo.
$lim_{x \to 0} sin x / x^{2}$ kyun exist nahi karta jabki $sin x / x^{3}$ deta hai $+ \infty$ ? ::: L'Hôpital ke baad denominators hain $2 x$ (sign flip hota hai) vs $3 x^{2}$ (hamesha positive).

Connections

L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms (index 4.1.26) — parent: rule aur uska proof.
Indeterminate Forms — matrix mein listed trouble ki saaton shapes.
Taylor Series — har "Verify" leading terms ke zariye yahan hai.
Exponential & Logarithm Growth Rates — Examples 3, 4, 6, 7 growth races hain.
Limits — Definition & Laws — one-sided limits aur $\pm \infty$ (Example 10).
Linear Approximation & Tangent Lines — $f^{'} / g^{'}$ sahi ratio kyun hai.