 for the numerator:
ln
⁡
(
1
+
4
𝑥
)
≈
4
𝑥
−
(
4
𝑥
)
2
2
+
𝑂
(
𝑥
3
)
=
4
𝑥
−
8
𝑥
2
+
𝑂
(
𝑥
3
)
ln(1+4x)≈4x− 
2
(4x) 
2
 

 +O(x 
3
 )=4x−8x 
2
 +O(x 
3
 )
𝑒
2
𝑥
−
1
≈
2
𝑥
+
(
2
𝑥
)
2
2
+
𝑂
(
𝑥
3
)
=
2
𝑥
+
2
𝑥
2
+
𝑂
(
𝑥
3
)
e 
2x
 −1≈2x+ 
2
(2x) 
2
 

 +O(x 
3
 )=2x+2x 
2
 +O(x 
3
 )
(
1
+
2
𝑥
)
3
−
1
=
1
+
6
𝑥
+
12
𝑥
2
+
8
𝑥
3
−
1
=
6
𝑥
+
12
𝑥
2
+
8
𝑥
3
(1+2x) 
3
 −1=1+6x+12x 
2
 +8x 
3
 −1=6x+12x 
2
 +8x 
3
 


Let's break down and verify the series expansions provided:

1. For \( \ln(1 + 4x) \):
   \[
   \ln(1 + 4x) \approx 4x - \frac{(4x)^2}{2} + O(x^3) = 4x - 8x^2 + O(x^3)
   \]
   This expansion is derived using the Taylor series expansion for \( \ln(1 + u) \), where \( u = 4x \):
   \[
   \ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots
   \]
   Substituting \( u = 4x \):
   \[
   \ln(1 + 4x) = 4x - \frac{(4x)^2}{2} + O(x^3) = 4x - 8x^2 + O(x^3)
   \]

2. For \( e^{2x} - 1 \):
   \[
   e^{2x} - 1 \approx 2x + \frac{(2x)^2}{2} + O(x^3) = 2x + 2x^2 + O(x^3)
   \]
   This expansion is derived using the Taylor series expansion for \( e^u \), where \( u = 2x \):
   \[
   e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots
   \]
   Substituting \( u = 2x \):
   \[
   e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + O(x^3) = 1 + 2x + 2x^2 + O(x^3)
   \]
   Subtracting 1 from both sides:
   \[
   e^{2x} - 1 = 2x + 2x^2 + O(x^3)
   \]

3. For \( (1 + 2x)^3 - 1 \):
   \[
   (1 + 2x)^3 - 1 = 1 + 6x + 12x^2 + 8x^3 - 1 = 6x + 12x^2 + 8x^3
   \]
   This expansion is derived using the binomial theorem for \( (1 + u)^n \), where \( u = 2x \) and \( n = 3 \):
   \[
   (1 + u)^n = 1 + \binom{n}{1}u + \binom{n}{2}u^2 + \binom{n}{3}u^3 + \cdots
   \]
   Substituting \( u = 2x \) and \( n = 3 \):
   \[
   (1 + 2x)^3 = 1 + 3 \cdot 2x + 3 \cdot (2x)^2 + (2x)^3 = 1 + 6x + 12x^2 + 8x^3
   \]
   Subtracting 1 from both sides:
   \[
   (1 + 2x)^3 - 1 = 6x + 12x^2 + 8x^3


Explore series expansions and approximations in calculus with detailed examples. Learn about Taylor series, binomial theorem, and their applications in mathematical analysis.

Math Problem Statement

for the numerator: ln ⁡ ( 1 + 4 𝑥 ) ≈ 4 𝑥 − ( 4 𝑥 ) 2 2 + 𝑂 ( 𝑥 3 )

+O(x 3 )=4x−8x 2 +O(x 3 ) 𝑒 2 𝑥 − 1 ≈ 2 𝑥 + ( 2 𝑥 ) 2 2 + 𝑂 ( 𝑥 3 )

+O(x 3 )=2x+2x 2 +O(x 3 ) ( 1 + 2 𝑥 ) 3 − 1

1 + 6 𝑥 + 12 𝑥 2 + 8 𝑥 3 − 1

Solution

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