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

True or false — justify

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

because

sin x

and

x

are equal near

0

False — they are never equal for

x \neq = 0

; their ratio approaches

1

because they shrink at the same rate. See Squeeze (Sandwich) Theorem.

lim_{x \to 0} \frac{s i n x}{x} = 1

holds whether

x

is in degrees or radians

False — only in radians. In degrees

sin x \approx \frac{π}{180} x

, so the limit is

\frac{π}{180} \approx 0.0175

. See Radian Measure.

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

equals

1^{\infty}

, which is

1

False —

1^{\infty}

is an indeterminate form; the base is

1 + tiny

, not exactly

1

, and the tiny excess compounds

n

times to reach

e

The proof of

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

using L'Hôpital's Rule is fully valid

False in spirit — L'Hôpital needs

\frac{d}{d x} sin x = cos x

, which itself is proved from this limit, so it is circular. The geometric squeeze avoids the circle.

lim_{x \to 0} \frac{s i n x}{x} = 1

means the sine graph is a straight line near the origin

Roughly true as a first approximation —

sin x \approx x

for small

x

, so the tangent at

0

has slope

1

; but

sin

still curves (next term

- \frac{x ^{3}}{6}

). See Taylor Series.

(1 + \frac{1}{n ^{2}})^{n} \to e

False — taking

ln

gives

n \cdot \frac{1}{n ^{2}} = \frac{1}{n} \to 0

, so the limit is

e^{0} = 1

. The base's decay must scale like

\frac{1}{n}

to hit

e

e

can be defined only as

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

False — equivalently

e = \sum_{k = 0}^{\infty} \frac{1}{k !}

, or as the base where

\frac{d}{d x} e^{x} = e^{x}

. All define the same number. See The Number e and ln.

lim_{x \to 0} \frac{1 - c o s x}{x} = \frac{1}{2}

False — that limit is

0

(odd/even powers matter). It is

\frac{1 - c o s x}{x ^{2}}

that gives

\frac{1}{2}

Spot the error

\frac{s i n x}{x} \to 0

since the top

sin x \to 0

" — find the flaw

The bottom also $\to 0$ , so it is

\frac{0}{0}

, not

\frac{0}{fixed}

. A

0/0

form measures relative speed, and here both shrink equally

\Rightarrow 1

lim (1 + \frac{1}{n})^{n} = (lim (1 + \frac{1}{n}))^{l i m n} = 1^{\infty} = 1

" — flaw?

You cannot split a limit of a variable base to a variable power like that; the base and exponent race against each other and must be evaluated together (log first).

\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?

No —

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

(same argument on top and bottom). The extra

\frac{5}{3}

came from the mismatched denominator

3 x

, not from the sine ratio.

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

because it's

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

" — flaw?

The bottom is

x

, not the argument

x^{2}

. Write

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

. The ratio must have the exact same expression top and bottom.

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

" — flaw?

Wrong log law.

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

, so it is

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

, a product, not

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

"Sector area

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

works with

x

in degrees" — flaw?

Only in radians. In degrees the sector area is

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

; the clean formula (and the whole proof) relies on Radian Measure.

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

, and both bounds

\to 1

, so we're done for all

x

" — flaw?

That chain was derived for

0 < x < \frac{π}{2}

. For

x < 0

you must also check the other side; since

\frac{s i n x}{x}

is even, the negative side gives the same limit, completing all cases.

Why questions

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

x = 0

Plugging in gives

\frac{0}{0}

, which is undefined; the squeeze traps the ratio between two things both heading to

1

, forcing the ratio to

1

without ever evaluating at

0

Why divide the inequality by

sin x

and not multiply by something?

Dividing by

sin x

isolates the target

\frac{s i n x}{x}

(after reciprocal); because

sin x > 0

0 < x < \frac{π}{2}

the inequality directions are preserved.

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

Reciprocals turn

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

into the useful

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

; taking reciprocals of positive quantities flips the inequality direction.

Why does the

e

-limit require taking a logarithm first?

The log converts the awkward exponent into a multiplier:

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

, turning a

1^{\infty}

fight into a manageable

0/0

limit. See Indeterminate Forms.

Why is

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

the natural fact to use for

e

It is the derivative of

ln

x = 1

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

. See The Number e and ln.

Why does

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

instead of

e

Log gives

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

, so the whole thing

\to e^{a}

. The constant

a

in the base sets the growth rate.

Why do radians make the arc length equal the angle?

By definition of Radian Measure, on a unit circle the arc subtended by angle

x

has length exactly

x

; that identity is what lets arc, chord, and

sin x

merge as

x \to 0

Edge cases

What is

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

(from the negative side)?

Still

1

— the function

\frac{s i n x}{x}

is even (both

sin

and

x

flip sign, cancelling), so left and right limits agree.

\frac{s i n x}{x}

defined at

x = 0

No — it is

\frac{0}{0}

there, an removable hole; the limit is

1

, so we can fill the hole by defining the value to be

1

(the sinc function).

What does

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

do for small

n

n = 1

n = 2

n = 1

gives

2

n = 2

gives

2.25

; the sequence increases toward

e \approx 2.718

but never reaches it — the limit is approached from below.

What is

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

(note the minus)?

It is

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

, using the master rule with

a = - 1

; a negative base-correction gives reciprocal growth.

What is

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

k = 0

0

— with

k = 0

the numerator is

sin 0 = 0

for all

x

, so the ratio is identically

0

; the formula "

\to k

" gives

0

consistently.

Does

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

hold as

x \to \frac{π}{2}

too?

No — near

\frac{π}{2}

tan x \to \infty

; the limit "

= 1

" is a local statement at

0

only, built from

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

and

cos x \to 1

What happens to

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

n

ranges over reals, not just integers?

Same limit

e

— the log argument

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

uses no integrality, so continuous

x \to \infty

works identically. See Derivative of sin and cos for the analogous continuous trick.

Connections

Squeeze (Sandwich) Theorem — the trap " $sin x / x \to 0$ " collapses once you remember the squeeze.
Radian Measure — root cause of the degrees trap.
Indeterminate Forms — every $1^{\infty}$ and $0/0$ trap here lives.
The Number e and ln — for the $e$ -definition traps.
L'Hôpital's Rule — the circularity trap.
Taylor Series — the " $sin$ is straight near $0$ " refinement.
Derivative of sin and cos — why the geometric proof matters.