Applications — approximation, evaluating limits: Question bank

Before the questions, a tiny glossary so every symbol below is earned:

True or false — justify

TF1. "A Taylor polynomial

P_{N}

equals

f

exactly."

False — it equals

f

plus the remainder

R_{N}

; only the full infinite series (inside its radius) equals

f

P_{N}

is an approximation, and

R_{N}

measures the gap.

TF2. "Adding more terms to a Taylor polynomial always makes the value at

x

more accurate."

Only inside the radius of convergence. Outside

R

the terms grow instead of shrink, so "more terms" makes it worse, not better.

TF3. "The Maclaurin series of

cos x

has only even powers because

cos

is an even function."

True — an even function's odd derivatives vanish at

0

, so every odd coefficient

c_{n} = f^{(n)} (0) / n!

0

, leaving only

1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

TF4. "The Lagrange remainder tells you the exact error."

False — it names an exact form

\frac{f ^{(N + 1)} ( c )}{( N + 1 )!} (x - a)^{N + 1}

, but

c

is unknown, so in practice you only get a bound by replacing

f^{(N + 1)} (c)

with its max

M

TF5. "For

sin (0.1)

, the error after

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

is at most

\frac{( 0.1 ) ^{5}}{120}

True —

sin

's Maclaurin series alternates with decreasing terms, so by the alternating-series bound the error is

\leq

the first omitted term

\frac{x ^{5}}{120}

TF6. "You may use the

ln (1 + x)

series at

x = 1

True at the endpoint — the interval is

- 1 < x \leq 1

, and

x = 1

converges (to

ln 2

) by the Alternating Series Test, even though

x = - 1

does not.

TF7. "A larger radius of convergence means a more accurate approximation for a fixed number of terms."

False — radius decides where the series works at all, not how fast it converges there. Accuracy at a point is governed by

∣ x - a ∣^{N + 1}

and

M

, not by

R

TF8. "Since the Maclaurin series of

e^{x}

converges for all

x

, using two terms is fine for

e^{10}

False — convergence is guaranteed eventually, but for large

x

you need many terms;

1 + 10 = 11

is nowhere near

e^{10} \approx 22026

Spot the error

SE1. "

lim_{x \to 0} \frac{s i n x - x}{x ^{3}}

: plug

x = 0

into

sin x = x - \frac{x ^{3}}{6} + \dots

giving

0 - 0 = 0

, so the limit is

0/0 = 0

Error — you plugged in before cancelling. Keep the series:

sin x - x = - \frac{x ^{3}}{6} + \dots

, divide by

x^{3}

, then let

x \to 0

to get

- \frac{1}{6}

SE2. "To find

lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}

, I expand

e^{x}

only to first order:

1 + x

. Then numerator

= 0

, limit

= 0

Error — expanding only to order

1

throws away the very term you need. The denominator is order

2

, so keep

e^{x}

through

\frac{x ^{2}}{2}

; the surviving leading term gives

\frac{1}{2}

SE3. "

x - tan x = - \frac{x ^{3}}{3}

, and

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

, so the ratio is

\frac{1}{6} \cdot \frac{1}{3} = \frac{1}{18}

Error — a ratio means divide, not multiply:

\frac{x ^{3} /6}{- x ^{3} /3} = \frac{1/6}{- 1/3} = - \frac{1}{2}

. Also the sign of

x - tan x

is negative.

SE4. "

ln (1 + x)

x = 2

equals

2 - \frac{4}{2} + \frac{8}{3} - \dots = ln 3

Error — the series diverges for

∣ x ∣ > 1

, so those terms add to nothing meaningful. To get

ln 3

you must recentre or use a convergent rearrangement; see Radius & Interval of Convergence.

SE5. "I bounded

e^{0.2}

's error with

M = 1

because

e^{c} \leq 1

[0, 0.2]

Error — on

[0, 0.2]

e^{c}

increases to

e^{0.2} \approx 1.22

, so

e^{c} \leq 1

is false. A safe bound is

M = 1.3

; a bound must be an upper limit on

∣ f^{(N + 1)} ∣

SE6. "For

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

I used

cos x \approx 1

, so numerator

= 0

, limit

= 0

Error — the

x^{2}

term of

cos

is exactly what survives:

1 - cos x = \frac{x ^{2}}{2} - \dots

, so dividing by

x^{2}

gives

\frac{1}{2}

. Always expand to the denominator's order.

SE7. "The alternating-series error bound applies to

e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots

Error — that series has all positive terms, so it does not alternate; the alternating bound is invalid. Use the Lagrange remainder

\frac{M}{( N + 1 )!} ∣ x ∣^{N + 1}

instead.

Why questions

WHY1. Why does substituting a series often beat L'Hôpital's Rule for

\frac{0}{0}

limits?

Series does all the "differentiating" once, up front, as coefficients; you cancel the lowest power in a single step instead of applying the quotient/derivative rule repeatedly.

WHY2. Why must you keep terms at least up to the denominator's degree in a limit?

The lower powers cancel to zero, so the answer is decided by the term whose order matches the denominator; drop it and you lose the entire limit.

WHY3. Why is the factorial

n!

in the denominator of a Taylor coefficient?

Differentiating

x^{n}

exactly

n

times produces

n!

; dividing by

n!

cancels it so that

c_{n}

reproduces

f

n

-th derivative and no more — see the derivation in Taylor & Maclaurin Series.

WHY4. Why can the leading (lowest-power) term dominate as

x \to 0

when it looks "smaller"?

For

∣ x ∣ < 1

, a smaller exponent gives a larger number (

x^{2} > x^{3}

), so the lowest power is the biggest; higher-order terms vanish faster and become negligible.

WHY5. Why does a smaller

∣ x - a ∣

make a fixed-order approximation better?

The error scales like

∣ x - a ∣^{N + 1}

; shrinking that distance shrinks the raised power dramatically, since raising a small number to a high power drives it toward zero.

WHY6. Why is

\frac{f ^{(N + 1)} ( c )}{( N + 1 )!} (x - a)^{N + 1}

the "next term in disguise"?

It has the exact shape of the

(N + 1)

-th Taylor term, but with the derivative evaluated at an unknown interior point

c

instead of at

a

— capturing everything the polynomial left out; see Lagrange Remainder & Error Estimation.

WHY7. Why can you add, multiply, and differentiate power series term by term inside the radius?

Inside

R

the series converges absolutely and uniformly enough that these operations behave like they do for polynomials — the justification lives in Power Series — Operations (add, multiply, differentiate).

WHY8. Why does the parent note call the polynomial the "best" fit at

a

Because its coefficients are the unique choice matching

f

's value and every derivative at

a

; any other polynomial disagrees with some slope or curvature there.

Edge cases

EC1. What happens to the series method for

lim_{x \to 0} \frac{t a n x - x}{x ^{3}}

if you only expand

tan x

to order

1

You get

0/ x^{3}

, an apparent

0

. You must expand

tan x = x + \frac{x ^{3}}{3} + \dots

to order

3

so the surviving

\frac{x ^{3}}{3}

gives the true limit

\frac{1}{3}

EC2. Is

x = 0

(the centre) a special case for a Maclaurin approximation?

Yes — at

x = 0

the error term

∣ x ∣^{N + 1} = 0

, so even

P_{0}

is exact: the polynomial reproduces

f (0)

perfectly with zero remainder.

EC3. What does the

ln (1 + x)

series do at its left endpoint

x = - 1

It becomes

- 1 - \frac{1}{2} - \frac{1}{3} - \dots

, the negative harmonic series, which diverges to

- \infty

— matching

ln (0)

. So

x = - 1

is excluded from the interval.

EC4. If

f

is a polynomial of degree

3

, what is its Taylor series about

a = 0

The polynomial itself — all derivatives past order

3

are zero, so

R_{3} = 0

and the "series" terminates; the approximation is exact everywhere.

EC5. Can two functions share the same first

N

Taylor terms yet differ?

Yes — e.g.

f (x) = 0

and

g (x) = e^{- 1/ x^{2}}

(with

g (0) = 0

) have identical zero Maclaurin coefficients but differ for

x \neq = 0

; matching finitely many terms never guarantees equality.

EC6. What is the limit

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

by series, and why is it not

\frac{0}{0}

trouble?

\frac{s i n x}{x} = \frac{x - x ^{3} /6 + \dots}{x} = 1 - \frac{x ^{2}}{6} + \dots \to 1

; the

x

in the numerator cancels the denominator cleanly, leaving the constant term

1

EC7. For an alternating series, when does the "error

\leq

first omitted term" rule fail?

When the terms are not monotonically decreasing in size; the bound requires each term smaller than the last. If they grow, use the Lagrange bound instead.

Recall One-line summary of the traps

Keep enough terms (denominator's order), respect the radius, bound before you trust, cancel by matching powers, and never confuse "the exact remainder form" with "a usable number".

Connections

Parent: Applications
Taylor & Maclaurin Series
Lagrange Remainder & Error Estimation
Alternating Series Test
Radius & Interval of Convergence
L'Hôpital's Rule
Power Series — Operations (add, multiply, differentiate)