Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e: Question bank

4.1.6 · D5 · Maths › Calculus I — Limits & Derivatives › Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

Tum ek variable base ko variable power wale limit ko aise split nahi kar sakte; base aur exponent ek doosre ke saath race karte hain aur inhe saath mein evaluate karna padta hai (pehle log lao).

\frac{s i n 5 x}{3 x} \to \frac{5}{3}

, so

\frac{s i n 5 x}{5 x} \to \frac{5}{3}

too" — flaw?

Nahi —

\frac{s i n 5 x}{5 x} \to 1

hota hai (top aur bottom par same argument). Extra

\frac{5}{3}

mismatched denominator

3 x

se aaya tha, sine ratio se nahi.

\frac{s i n ( x ^{2} )}{x} \to 1

because it's

\frac{s i n ( stuff )}{stuff}

" — flaw?

Bottom

x

hai, argument

x^{2}

nahi. Likho

\frac{s i n ( x ^{2} )}{x ^{2}} \cdot x \to 1 \cdot 0 = 0

. Ratio mein top aur bottom mein exactly same expression hona chahiye.

ln (1 + \frac{1}{n})^{n} = ln (1 + \frac{1}{n}) \cdot ln n

" — flaw?

Galat log law hai.

ln (a^{b}) = b ln a

, toh yeh

n ln (1 + \frac{1}{n})

hai, ek product, na ki

ln (1 + \frac{1}{n}) \cdot ln n

"Sector area

= \frac{1}{2} r^{2} x

works with

x

in degrees" — flaw?

Sirf radians mein. Degrees mein sector area

\frac{1}{2} r^{2} \cdot \frac{π x}{180}

hai; clean formula (aur poora proof) Radian Measure par depend karta hai.

cos x \leq \frac{s i n x}{x} \leq 1

, and both bounds

\to 1

, so we're done for all

x

" — flaw?

Yeh chain

0 < x < \frac{π}{2}

ke liye derive ki gayi thi.

x < 0

ke liye tumhe doosri side bhi check karni hogi; kyunki

\frac{s i n x}{x}

even hai, negative side same limit deti hai, jisse sab cases complete hote hain.

Why questions

Why must we use the Squeeze (Sandwich) Theorem and not just plug in

x = 0

Plug in karne se

\frac{0}{0}

milta hai, jo undefined hai; squeeze ratio ko do cheezون ke beech trap karta hai jo dono

1

ki taraf ja rahe hain, ratio ko

1

force karta hai bina

0

par evaluate kiye.

Why divide the inequality by

sin x

and not multiply by something?

sin x

se divide karne par target

\frac{s i n x}{x}

isolate hota hai (reciprocal ke baad); kyunki

sin x > 0

hai

0 < x < \frac{π}{2}

par, inequality directions preserve rehti hain.

Why take reciprocals in the middle of the proof, and what does it do to the inequalities?

Reciprocals

1 \leq \frac{x}{s i n x} \leq \frac{1}{c o s x}

ko useful

cos x \leq \frac{s i n x}{x} \leq 1

mein convert karte hain; positive quantities ke reciprocals lena inequality direction flip kar deta hai.

Why does the

e

-limit require taking a logarithm first?

Log awkward exponent ko multiplier mein convert karta hai:

ln (1 + \frac{1}{n})^{n} = n ln (1 + \frac{1}{n})

, ek

1^{\infty}

fight ko manageable

0/0

limit mein badal deta hai. Dekho Indeterminate Forms.

Why is

lim_{h \to 0} \frac{l n ( 1 + h )}{h} = 1

the natural fact to use for

e

Yeh

x = 1

par

ln

ki derivative hi hai:

\frac{l n ( 1 + h ) - l n 1}{h} \to \frac{d}{d x} ln x_{1} = 1

. Dekho The Number e and ln.

Why does

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

instead of

e

Log deta hai

x ln (1 + \frac{a}{x}) = \frac{l n ( 1 + a / x )}{1/ x} \to a

, toh poori cheez

\to e^{a}

. Base mein constant

a

growth rate set karta hai.

Why do radians make the arc length equal the angle?

Radian Measure ki definition se, unit circle par angle

x

se subtended arc ki length exactly

x

hoti hai; yahi identity hai jo arc, chord, aur

sin x

x \to 0

par merge karne deti hai.

Edge cases

What is

lim_{x \to 0^{-}} \frac{s i n x}{x}

(from the negative side)?

Phir bhi

1

— function

\frac{s i n x}{x}

even hai (dono

sin

aur

x

sign flip karte hain, cancel hoke), toh left aur right limits agree karte hain.

\frac{s i n x}{x}

defined at

x = 0

Nahi — wahan yeh

\frac{0}{0}

hai, ek removable hole; limit

1

hai, toh hum hole fill kar sakte hain value ko

1

define karke (sinc function).

What does

(1 + \frac{1}{n})^{n}

do for small

n

n = 1

n = 2

n = 1

deta hai

2

n = 2

deta hai

2.25

; sequence

e \approx 2.718

ki taraf increase karta hai lekin kabhi pahunchta nahi — limit neeche se approach hoti hai.

What is

lim_{x \to \infty} (1 - \frac{1}{x})^{x}

(note the minus)?

Yeh

e^{- 1} = \frac{1}{e}

hai, master rule ko

a = - 1

ke saath use karke; negative base-correction reciprocal growth deta hai.

What is

lim_{x \to 0} \frac{s i n ( k x )}{x}

k = 0

0

—

k = 0

ke saath numerator

sin 0 = 0

hai sab

x

ke liye, toh ratio identically

0

hai; formula "

\to k

" consistently

0

deta hai.

Does

lim_{x \to 0} \frac{t a n x}{x} = 1

hold as

x \to \frac{π}{2}

too?

Nahi —

\frac{π}{2}

ke paas,

tan x \to \infty

; limit "

= 1

" sirf

0

par ek local statement hai, jo

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

aur

cos x \to 1

se banta hai.

What happens to

(1 + \frac{1}{n})^{n}

n

ranges over reals, not just integers?

Same limit

e

— log argument

\frac{l n ( 1 + 1/ x )}{1/ x} \to 1

mein koi integrality use nahi hoti, toh continuous

x \to \infty

bilkul same kaam karta hai. Dekho Derivative of sin and cos analogous continuous trick ke liye.

Connections

Squeeze (Sandwich) Theorem — trap " $sin x / x \to 0$ " collapse ho jaata hai jab tum squeeze yaad karte ho.
Radian Measure — degrees trap ki root cause.
Indeterminate Forms — yahaan har $1^{\infty}$ aur $0/0$ trap rehta hai.
The Number e and ln — $e$ -definition traps ke liye.
L'Hôpital's Rule — circularity trap.
Taylor Series — " $sin$ $0$ ke paas straight hai" refinement.
Derivative of sin and cos — kyun geometric proof matter karta hai.

Question bank — Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

True or false — justify

Spot the error

Why questions

Edge cases

Connections