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

4.1.26 · D5 · Maths › Calculus I — Limits & Derivatives › L'Hôpital's rule — proof using linear approximation, 0 - 0,

True or false — justify

Jab bhi koi limit

\frac{0}{0}

deta hai, L'Hôpital's rule answer de degi.

False — yeh fail bhi ho sakti hai agar

lim f^{'} / g^{'}

exist na kare ya loop karta rahe; phir algebra, Taylor, ya Limits — Definition & Laws pe wapas jaana padta hai.

\frac{x + s i n x}{x}

waala trap dekho.

\frac{0}{0}

ek number hai jo

1

ke barabar hai kyunki koi bhi cheez apne aap se divide ho toh

1

hoti hai.

False —

\frac{0}{0}

ek indeterminate form hai, ek sawaal hai na ki koi value;

\frac{s i n x}{x} \to 1

lekin

\frac{s i n x}{x ^{2}} \to \infty

, dono

\frac{0}{0}

jaise dikhte hain. Indeterminate Forms dekho.

L'Hôpital fraction

\frac{f}{g}

ko quotient rule se differentiate karta hai.

False — yeh numerator aur denominator ko alag-alag differentiate karta hai, jisse

\frac{f ^{'}}{g ^{'}}

milta hai, na ki

\frac{f ^{'} g - f g ^{'}}{g ^{2}}

. Yeh dono bilkul alag operations hain.

Agar

lim \frac{f ^{'}}{g ^{'}}

exist nahi karta, toh

lim \frac{f}{g}

bhi exist nahi karta.

False — rule ek direction mein hi kaam karta hai.

\frac{x + s i n x}{x} \to 1

lekin

\frac{1 + c o s x}{1}

oscillate karta hai aur koi limit nahi hai; RHS ka fail hona kuch prove nahi karta.

Tum L'Hôpital ko

\frac{0}{5}

pe apply kar sakte ho kyunki upar

0

hai.

False — sirf

\frac{0}{0}

aur

\frac{\infty}{\infty}

qualify karte hain.

\frac{0}{5} = 0

seedha hai; yahan rule force karne se galat ya meaningless answer aata hai.

0 \cdot \infty

ko seedha L'Hôpital mein daala ja sakta hai.

False — rule ko ek fraction chahiye. Pehle

f \cdot g = \frac{f}{1/ g}

likhkar

\frac{0}{0}

\frac{\infty}{\infty}

banao.

1^{\infty}

barabar

1

hai kyunki

1

ki koi bhi power

1

hoti hai.

False — base sirf

1

ki taraf jaata hai jabki exponent badhta rehta hai;

(1 + \frac{1}{x})^{x} \to e

, na ki

1

. Isko resolve karne ke liye

ln

lo.

Pehli application ke baad L'Hôpital ko dobara lagana hamesha legal hai.

False — har reapplication se pehle dobara check karna zaroori hai ki naya fraction abhi bhi

\frac{0}{0}

\frac{\infty}{\infty}

hai ya nahi. Ise non-indeterminate form pe lagana ek galti hai.

Linear-approximation derivation ek bilkul rigorous proof hai.

False — yeh secretly assume karta hai ki

f^{'}, g^{'}

point

a

pe continuous hain. Airtight proof Cauchy Mean Value Theorem use karta hai, jisme aisa koi assumption nahi hota.

L'Hôpital one-sided limits aur

x \to \pm \infty

ke liye kaam karta hai.

True — theorem

a

finite,

a \to \pm \infty

, aur one-sided limits ke liye stated hai, bas relevant side pe hypotheses hold karne chahiye.

Why questions

Linear-approximation derivation mein constant terms kyun vanish ho jaate hain?

Kyunki

f (a) = g (a) = 0

hai (

\frac{0}{0}

case mein continuity guarantee karta hai), isliye tangent lines height zero se guzarti hain aur sirf slope terms bachte hain. Linear Approximation & Tangent Lines dekho.

Shared factor

(x - a)

cancel kyun hota hai, aur yeh kya represent karta hai?

Dono functions point ki taraf same rate

(x - a)

se shrink karte hain; woh shared "smallness" cancel ho jaati hai, aur slopes ka ratio

\frac{f ^{'} ( a )}{g ^{'} ( a )}

— yaani speeds ka ratio — bach jaata hai.

0^{0}

\infty^{0}

1^{\infty}

handle karne ke liye

ln

kyun lena padta hai?

Variable exponent differentiation se invisible rehta hai jab tak

ln

use karke use product ke roop mein neeche nahi laate:

ln (f^{g}) = g ln f

, jo

0 \cdot \infty

mein convert ho jaata hai aur phir hum usse fraction bana sakte hain.

Rule ke liye right-hand limit ka exist karna zaroori kyun hai?

Kyunki equality sirf yeh guarantee karti hai ki "agar RHS exist kare toh LHS uske barabar hai." Non-existent RHS bridge tod deta hai — LHS dusre tareekon se abhi bhi exist kar sakta hai. Exponential & Logarithm Growth Rates dekho un cases ke liye jo algebra maangti hain.

Cauchy MVT tangent-line picture se zyada strong kyun hai?

Yeh ek exact intermediate point

c

produce karta hai jahan

\frac{f ( x )}{g ( x )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}

sirf differentiability use karke — koi assumption nahi ki

f^{'}, g^{'}

continuous hain. Mean Value Theorem dekho.

\frac{l n x}{x} \to 0

growth rates ke baare mein kya bata hai?

Yeh dikhata hai ki

ln x

x

ki kisi bhi positive power se slower grow karta hai — logarithm polynomials ke against har race haarta hai. Exponential & Logarithm Growth Rates dekho.

Taylor series kaafi limits ke liye L'Hôpital ko replace kyun kar sakti hai?

Limit ka decision upar aur neeche ke leading nonzero terms se hota hai; unka ratio answer deta hai, aksar repeated differentiation se bhi tez. Taylor Series dekho.

Edge cases

Agar tum L'Hôpital ko

\frac{\infty}{- \infty}

pe apply karo toh kya hoga?

Legal hai — rule kisi bhi

\frac{\pm \infty}{\pm \infty}

ko cover karta hai; sign bas final answer mein carry through hoti hai.

Kya L'Hôpital

\infty - \infty

ko seedha handle kar sakta hai?

Nahi — pehle common denominator pe combine karo ya factor karo taaki

\frac{0}{0}

bane, phir rule apply karo.

Agar

g^{'} (x) = 0

point

a

ke arbitrarily close ho toh kya hoga?

Hypothesis "

g^{'} (x) \neq = 0

near

a

" fail ho jaati hai, isliye theorem apply nahi hota; har neighbourhood mein

\frac{f ^{'}}{g ^{'}}

legitimately form nahi ho sakta.

Agar

f

aur

g

dono polynomials hain jo

a

pe vanish karte hain, toh kya L'Hôpital zaroori hai?

Nahi —

(x - a)

factor out karo aur cancel karo. L'Hôpital kaam karta hai lekin factoring cleaner hai aur directly multiplicity reveal karta hai.

Kya rule tab bhi kaam karta hai jab limit

\frac{0}{0}

ho lekin derivatives

a

pe exist na karein?

Haan — differentiability

a

ke paas chahiye (zaruri nahi

a

pe), jo Cauchy-MVT proof se match karta hai jisme sirf ek interior point

c \to a

sample hota hai.

Agar rule lagane se

\frac{0}{0}

ek determinate form jaise

\frac{3}{0 ^{+}}

mein badal jaaye toh kya karein?

Ruko — naya form ab indeterminate nahi raha;

+ \infty

padh lo. Dobara apply karna galat hoga kyunki form

\frac{0}{0}

\frac{\infty}{\infty}

nahi hai.

Connections

Parent: L'Hôpital's Rule
Mean Value Theorem — traps ke peeche ka rigor.
Indeterminate Forms — kyun $\frac{0}{0}$ ek sawaal hai.
Taylor Series — alternative jab L'Hôpital loop kare.
Exponential & Logarithm Growth Rates — growth-race intuition.
Limits — Definition & Laws — ground rules.

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

True or false — justify

Spot the error

Why questions

Edge cases

Connections