Single-Variable-Calculus · Unit 7 · Video 2 · Interactive Practice

The Standard Linear Approximations Near Zero: sin x, cos x, e^x, ln(1+x), (1+x)^r

IKey Formulas

Formula	Name	Where it comes from
$f(x) \approx f(0) + f'(0)\,x \quad (x \approx 0)$	Linear approximation at $0$	Tangent-line formula with $x_0 = 0$
$\sin x \approx x, \qquad \cos x \approx 1$	Sine and cosine	$\sin 0 = 0,\ \cos 0 = 1$ ; then $\cos 0 = 1,\ -\sin 0 = 0$
$e^x \approx 1 + x$	Exponential	$f(0) = f'(0) = e^0 = 1$
$\ln(1+x) \approx x, \qquad (1+x)^r \approx 1 + rx$	Log and power	$f(0) = 0,\ f'(0) = 1$ ; then $f(0) = 1,\ f'(0) = r$

Key Insight: In every formula the constant term is $f(0)$ and the coefficient of $x$ is $f'(0)$ . The left sides need a calculator; the right sides are linear in $x$ and need none.

IIVisualization 1 — Five Tangent Lines at Zero

Each approximation is the tangent line at $x = 0$ : height $f(0)$ , slope $f'(0)$ .

f(x) r 1/2 x 0.50

💡 The list uses $1 + x$ because $0$ is a bad base point for $\ln x$ (it falls to $-\infty$ ) and for $\sqrt{x}$ (its slope $\tfrac{1}{2\sqrt{x}} \to \infty$ ): expand at $1$ , then shift, as in $\ln u \approx u - 1$ with $u = 1 + x$ .

IIIVisualization 2 — Zooming In on $x = 0$

Near $0$ the gap between curve and tangent line shrinks faster than the window, so the two merge.

f(x) zoom ±0.9

IVVisualization 3 — Estimates at a Glance

With $x$ near $0$ , one line of the list turns a calculator value into mental arithmetic.

Estimate your estimate 0.000

VQuiz Questions

Problem 1 · A Power Near One

Given: $(1+x)^r \approx 1 + rx$ for $x \approx 0$ — estimate $(1.02)^5$ .

✅ Correct! With

x = 0.02

and

r = 5

1 + 5(0.02) = 1.10

. (The exact value is

1.10408\ldots

)

❌ Missing the factor $r$ . The slope at

0

f'(0) = r = 5

, so the linear term is

5x

, not

x

❌ That is $r(1+x)$ . The approximation is

1 + rx

: the constant term is

f(0) = 1

and only

x

gets multiplied by

r

❌ That is $1 + x/r$ . The coefficient of

x

f'(0) = r

, so multiply by

5

, don't divide.

❌ Not quite. Write

1.02 = 1 + 0.02

and use

1 + rx

Show solution

Write $1.02 = 1 + x$ with $x = 0.02$ , near $0$ . For $f(x) = (1+x)^5$ : $f(0) = 1$ and $f'(x) = 5(1+x)^4$ , so $f'(0) = 5$ .

(1.02)^5 \approx 1 + 5(0.02) = 1 + 0.10 = 1.10

A calculator gives $1.10408\ldots$ , so the estimate is off by about $0.004$ .

Problem 2 · Reading Off x Correctly

Given: $\ln(1+x) \approx x$ for $x \approx 0$ — estimate $\ln(0.97)$ .

✅ Correct!

0.97 = 1 + (-0.03)

, so

x = -0.03

and

\ln(0.97) \approx -0.03

(exact:

-0.03046\ldots

❌ $x$ is not $0.97$ . In

\ln(1+x)

x

is how far the argument is from

1

0.97 = 1 + x

gives

x = -0.03

. Using

0.97

itself would need

x

near

0

, and

0.97

is not.

❌ Check the sign.

0.97

is below

1

, so

x = 0.97 - 1 = -0.03

, and

\ln

of a number below

1

is negative.

❌ That keeps only $f(0) = \ln 1 = 0$ . The slope term

f'(0)\,x = 1 \cdot x

is the whole estimate here.

❌ Not quite. Solve

1 + x = 0.97

for

x

first.

Show solution

Match $0.97$ to $1 + x$ : $x = 0.97 - 1 = -0.03$ , which is near $0$ . Then

\ln(0.97) = \ln\big(1 + (-0.03)\big) \approx -0.03.

Equivalently, the last video's $\ln u \approx u - 1$ near $u = 1$ gives $0.97 - 1 = -0.03$ — the same computation after $u = 1 + x$ . A calculator gives $-0.03046\ldots$

Problem 3 · A Reciprocal Square Root

Given: $\dfrac{1}{\sqrt{1.04}}$ — write it as $(1+x)^r$ and estimate it with $(1+x)^r \approx 1 + rx$ .

Which x and r?

What is the estimate?

✅ Correct!

(1 + 0.04)^{-1/2} \approx 1 - \tfrac{1}{2}(0.04) = 0.98

; the exact value is

0.98058\ldots

❌ Check the sign of $r$ .

\sqrt{1.04} = (1.04)^{1/2}

, and the reciprocal flips the exponent:

(1.04)^{-1/2}

❌ $x$ is the offset from $1$ .

1.04 = 1 + 0.04

, so

x = 0.04

;

x = 1.04

is not near

0

❌ A square root is the power $\tfrac{1}{2}$ , not $2$ . So

\tfrac{1}{\sqrt{1.04}} = (1.04)^{-1/2}

❌ Not quite. Write

1.04 = 1 + x

, and write the reciprocal square root as a single power.

❌ That is $\sqrt{1.04} \approx 1.02$ . With

r = -\tfrac{1}{2}

the correction is negative:

1 - 0.02

❌ The factor $r$ is $-\tfrac{1}{2}$ , not $-1$ .

rx = -\tfrac{1}{2}(0.04) = -0.02

❌ That uses $r = -2$ . With

r = -\tfrac{1}{2}

1 + rx = 1 - 0.02

❌ Not quite. Compute

1 + rx

with

r = -\tfrac{1}{2}

and

x = 0.04

Show solution

Step 1 — Rewrite as a power: $\dfrac{1}{\sqrt{1.04}} = (1.04)^{-1/2} = (1 + 0.04)^{-1/2}$ , so $x = 0.04$ and $r = -\tfrac{1}{2}$ .

Step 2 — Apply the formula ( $x = 0.04$ is near $0$ ):

(1 + 0.04)^{-1/2} \approx 1 + \left(-\tfrac{1}{2}\right)(0.04) = 1 - 0.02 = 0.98

A calculator gives $0.98058\ldots$

Problem 4 · Factor First, Then Approximate

Given: $\sqrt{4.08}$ — estimate it using a linear approximation from the list.

✅ Correct!

\sqrt{4.08} = 2\sqrt{1.02} \approx 2\left(1 + \tfrac{1}{2}(0.02)\right) = 2.02

; the exact value is

2.01990\ldots

❌ Missing $r = \tfrac{1}{2}$ .

\sqrt{1.02} \approx 1 + \tfrac{1}{2}(0.02) = 1.01

, not

1.02

❌ $x = 3.08$ is not near $0$ . Writing

4.08 = 1 + 3.08

gives

1 + \tfrac{1}{2}(3.08) = 2.54

, far from the truth; factor out

4

so that

x

is small.

❌ That is $\sqrt{4} + 0.08$ . The square root shrinks the extra

0.08

: factor out

4

and use

(1+x)^{1/2} \approx 1 + \tfrac{1}{2}x

❌ Not quite. Factor

4.08 = 4(1.02)

first.

Show solution

Step 1 — Factor so the argument is near $1$ : $4.08 = 4 \cdot 1.02$ , so $\sqrt{4.08} = \sqrt{4}\,\sqrt{1.02} = 2\,(1 + 0.02)^{1/2}$ .

Step 2 — Apply $(1+x)^r \approx 1 + rx$ with $x = 0.02$ , $r = \tfrac{1}{2}$ :

\sqrt{4.08} \approx 2\left(1 + \tfrac{1}{2}(0.02)\right) = 2(1.01) = 2.02

A calculator gives $2.01990\ldots$ . Without factoring, $x = 3.08$ would be far from $0$ and the estimate $2.54$ would be useless.

Solved: 0 / 4

The Standard Linear Approximations Near Zero: sin x, cos x, e^x, ln(1+x), (1+x)^r

IKey Formulas

IIVisualization 1 — Five Tangent Lines at Zero

IIIVisualization 2 — Zooming In on x=0x = 0x=0

IVVisualization 3 — Estimates at a Glance

VQuiz Questions

IIIVisualization 2 — Zooming In on $x = 0$