Comprehensive worksheet featuring a wide variety of limit problems ranging from basic algebraic expressions to complex trigonometric and exponential functions.

Problem: Solve the given limits and explain the solution for each.

We will solve each limit step by step using techniques such as L'Hôpital's Rule, Taylor series expansions, and algebraic simplifications where applicable.

---

#### a) \( \lim_{x \to 1} \frac{x^3 - 2x^2 + x}{x^2 - 1} \)

Solution:
First, factor the numerator and the denominator:
- Numerator: \( x^3 - 2x^2 + x = x(x^2 - 2x + 1) = x(x-1)^2 \)
- Denominator: \( x^2 - 1 = (x-1)(x+1) \)

Thus, the expression becomes:
\[
\lim_{x \to 1} \frac{x(x-1)^2}{(x-1)(x+1)} = \lim_{x \to 1} \frac{x(x-1)}{x+1}
\]

Now, substitute \( x = 1 \):
\[
\frac{1(1-1)}{1+1} = \frac{0}{2} = 0
\]

Answer:
\[
\boxed{0}
\]

---

#### b) \( \lim_{x \to 3} \frac{x^3 - 9x}{x^4 - 3x^3 - x + 3} \)

Solution:
Factor the numerator and the denominator:
- Numerator: \( x^3 - 9x = x(x^2 - 9) = x(x-3)(x+3) \)
- Denominator: \( x^4 - 3x^3 - x + 3 \). We test \( x = 3 \) as a root:
\[
3^4 - 3 \cdot 3^3 - 3 + 3 = 81 - 81 - 3 + 3 = 0
\]
So, \( x-3 \) is a factor. Perform polynomial division or factorization to get:
\[
x^4 - 3x^3 - x + 3 = (x-3)(x^3 - x - 1)
\]

Thus, the expression becomes:
\[
\lim_{x \to 3} \frac{x(x-3)(x+3)}{(x-3)(x^3 - x - 1)} = \lim_{x \to 3} \frac{x(x+3)}{x^3 - x - 1}
\]

Now, substitute \( x = 3 \):
\[
\frac{3(3+3)}{3^3 - 3 - 1} = \frac{3 \cdot 6}{27 - 3 - 1} = \frac{18}{23}
\]

Answer:
\[
\boxed{\frac{18}{23}}
\]

---

#### c) \( \lim_{x \to 1} \frac{\sqrt[3]{8x} - 2x}{\sqrt[4]{x} - x} \)

Solution:
Let \( y = \sqrt[3]{8x} \). Then \( y^3 = 8x \) and \( x = \frac{y^3}{8} \). As \( x \to 1 \), \( y \to 2 \). The limit becomes:
\[
\lim_{y \to 2} \frac{y - 2 \cdot \frac{y^3}{8}}{\sqrt[4]{\frac{y^3}{8}} - \frac{y^3}{8}}
\]

This is complicated, so we use L'Hôpital's Rule directly on the original form:
\[
\lim_{x \to 1} \frac{\frac{d}{dx}(\sqrt[3]{8x} - 2x)}{\frac{d}{dx}(\sqrt[4]{x} - x)} = \lim_{x \to 1} \frac{\frac{8}{3}(8x)^{-2/3} - 2}{\frac{1}{4}x^{-3/4} - 1}
\]

Evaluate at \( x = 1 \):
\[
\frac{\frac{8}{3}(8 \cdot 1)^{-2/3} - 2}{\frac{1}{4}(1)^{-3/4} - 1} = \frac{\frac{8}{3} \cdot \frac{1}{4} - 2}{\frac{1}{4} - 1} = \frac{\frac{2}{3} - 2}{\frac{1}{4} - 1} = \frac{\frac{2}{3} - \frac{6}{3}}{\frac{1}{4} - \frac{4}{4}} = \frac{-\frac{4}{3}}{-\frac{3}{4}} = \frac{4}{3} \cdot \frac{4}{3} = \frac{16}{9}
\]

Answer:
\[
\boxed{\frac{16}{9}}
\]

---

#### d) \( \lim_{x \to 1} \frac{\sqrt{x+1} - \sqrt{2}}{x^2 - 1} \)

Solution:
Rationalize the numerator:
\[
\lim_{x \to 1} \frac{\sqrt{x+1} - \sqrt{2}}{x^2 - 1} \cdot \frac{\sqrt{x+1} + \sqrt{2}}{\sqrt{x+1} + \sqrt{2}} = \lim_{x \to 1} \frac{(x+1) - 2}{(x^2 - 1)(\sqrt{x+1} + \sqrt{2})}
\]
\[
= \lim_{x \to 1} \frac{x - 1}{(x-1)(x+1)(\sqrt{x+1} + \sqrt{2})} = \lim_{x \to 1} \frac{1}{(x+1)(\sqrt{x+1} + \sqrt{2})}
\]

Substitute \( x = 1 \):
\[
\frac{1}{(1+1)(\sqrt{1+1} + \sqrt{2})} = \frac{1}{2(2\sqrt{2})} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}
\]

Answer:
\[
\boxed{\frac{\sqrt{2}}{8}}
\]

---

#### e) \( \lim_{x \to 0} \frac{5^x - 1}{x} \)

Solution:
Use the definition of the derivative or the known limit \( \lim_{x \to 0} \frac{a^x - 1}{x} = \ln(a) \):
\[
\lim_{x \to 0} \frac{5^x - 1}{x} = \ln(5)
\]

Answer:
\[
\boxed{\ln(5)}
\]

---

#### f) \( \lim_{x \to 1} \frac{3x^3 - 3}{3^x - 3} \)

Solution:
Factor the numerator:
\[
\lim_{x \to 1} \frac{3(x^3 - 1)}{3^x - 3} = \lim_{x \to 1} \frac{3(x-1)(x^2 + x + 1)}{3^x - 3}
\]

Use L'Hôpital's Rule:
\[
\lim_{x \to 1} \frac{3(3x^2 + 2x + 1)}{3^x \ln(3)} = \frac{3(3 \cdot 1^2 + 2 \cdot 1 + 1)}{3^1 \ln(3)} = \frac{3(3 + 2 + 1)}{3 \ln(3)} = \frac{3 \cdot 6}{3 \ln(3)} = \frac{6}{\ln(3)}
\]

Answer:
\[
\boxed{\frac{6}{\ln(3)}}
\]

---

#### g) \( \lim_{x \to 0} \frac{e^x - e^{-x}}{x} \)

Solution:
Use the definition of the derivative or expand using Taylor series:
\[
e^x = 1 + x + \frac{x^2}{2!} + \cdots, \quad e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots
\]
\[
e^x - e^{-x} = (1 + x + \frac{x^2}{2!} + \cdots) - (1 - x + \frac{x^2}{2!} - \cdots) = 2x + \frac{2x^3}{3!} + \cdots
\]
\[
\frac{e^x - e^{-x}}{x} = \frac{2x + \frac{2x^3}{3!} + \cdots}{x} = 2 + \frac{2x^2}{3!} + \cdots
\]

As \( x \to 0 \):
\[
\lim_{x \to 0} \frac{e^x - e^{-x}}{x} = 2
\]

Answer:
\[
\boxed{2}
\]

---

#### h) \( \lim_{x \to 0} \frac{e^x - 1}{x^2} \)

Solution:
Use L'Hôpital's Rule:
\[
\lim_{x \to 0} \frac{e^x - 1}{x^2} = \lim_{x \to 0} \frac{e^x}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}
\]

Answer:
\[
\boxed{\infty}
\]

---

#### i) \( \lim_{x \to \frac{\pi}{3}} \frac{1 - 2\cos x}{\pi - 3x} \)

Solution:
Use L'Hôpital's Rule:
\[
\lim_{x \to \frac{\pi}{3}} \frac{1 - 2\cos x}{\pi - 3x} = \lim_{x \to \frac{\pi}{3}} \frac{2\sin x}{-3} = \frac{2\sin\left(\frac{\pi}{3}\right)}{-3} = \frac{2 \cdot \frac{\sqrt{3}}{2}}{-3} = \frac{\sqrt{3}}{-3} = -\frac{\sqrt{3}}{3}
\]

Answer:
\[
\boxed{-\frac{\sqrt{3}}{3}}
\]

---

#### j) \( \lim_{x \to 0} \frac{\sin x}{e^x - 1} \)

Solution:
Use L'Hôpital's Rule:
\[
\lim_{x \to 0} \frac{\sin x}{e^x - 1} = \lim_{x \to 0} \frac{\cos x}{e^x} = \frac{\cos(0)}{e^0} = \frac{1}{1} = 1
\]

Answer:
\[
\boxed{1}
\]

---

#### k) \( \lim_{x \to 0} \frac{\arctan 3x}{\arcsin 2x} \)

Solution:
Use the small-angle approximations \( \arctan(3x) \approx 3x \) and \( \arcsin(2x) \approx 2x \):
\[
\lim_{x \to 0} \frac{\arctan 3x}{\arcsin 2x} = \lim_{x \to 0} \frac{3x}{2x} = \frac{3}{2}
\]

Answer:
\[
\boxed{\frac{3}{2}}
\]

---

#### l) \( \lim_{x \to 0} \frac{e^x - e^{-x}}{2x - \sin x} \)

Solution:
Use Taylor series expansions:
\[
e^x = 1 + x + \frac{x^2}{2!} + \cdots, \quad e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots
\]
\[
e^x - e^{-x} = 2x + \frac{2x^3}{3!} + \cdots
\]
\[
2x - \sin x = 2x - \left(x - \frac{x^3}{3!} + \cdots\right) = x + \frac{x^3}{6} + \cdots
\]
\[
\frac{e^x - e^{-x}}{2x - \sin x} = \frac{2x + \frac{2x^3}{3!} + \cdots}{x + \frac{x^3}{6} + \cdots} = \frac{2 + \frac{2x^2}{3!} + \cdots}{1 + \frac{x^2}{6} + \cdots}
\]

As \( x \to 0 \):
\[
\lim_{x \to 0} \frac{e^x - e^{-x}}{2x - \sin x} = 2
\]

Answer:
\[
\boxed{2}
\]

---

#### m) \( \lim_{x \to 0} \frac{x^3 + \pi x}{\sin 3x} \)

Solution:
Use the small-angle approximation \( \sin(3x) \approx 3x \):
\[
\lim_{x \to 0} \frac{x^3 + \pi x}{\sin 3x} = \lim_{x \to 0} \frac{x^3 + \pi x}{3x} = \lim_{x \to 0} \frac{x^2 + \pi}{3} = \frac{\pi}{3}
\]

Answer:
\[
\boxed{\frac{\pi}{3}}
\]

---

#### n) \( \lim_{x \to 0} \frac{\ln(1 + 4x)}{3^x - 1} \)

Solution:
Use L'Hôpital's Rule:
\[
\lim_{x \to 0} \frac{\ln(1 + 4x)}{3^x - 1} = \lim_{x \to 0} \frac{\frac{4}{1 + 4x}}{3^x \ln(3)} = \frac{4}{1} \cdot \frac{1}{\ln(3)} = \frac{4}{\ln(3)}
\]

Answer:
\[
\boxed{\frac{4}{\ln(3)}}
\]

---

#### o) \( \lim_{x \to 0} \frac{\sin 4x}{\ln(1 + \sin x)} \)

Solution:
Use the small-angle approximations \( \sin(4x) \approx 4x \) and \( \ln(1 + \sin x) \approx \sin x \approx x \):
\[
\lim_{x \to 0} \frac{\sin 4x}{\ln(1 + \sin x)} = \lim_{x \to 0} \frac{4x}{x} = 4
\]

Answer:
\[
\boxed{4}
\]

---

#### p) \( \lim_{x \to 0} \frac{3\ln(1 - 2x)}{2\arctan 3x} \)

Solution:
Use the small-angle approximations \( \ln(1 - 2x) \approx -2x \) and \( \arctan(3x) \approx 3x \):
\[
\lim_{x \to 0} \frac{3\ln(1 - 2x)}{2\arctan 3x} = \lim_{x \to 0} \frac{3(-2x)}{2(3x)} = \frac{-6x}{6x} = -1
\]

Answer:
\[
\boxed{-1}
\]

---

#### q) \( \lim_{x \to 0} \frac{\arctan x + x^2}{2^{3x} - 3^{2x}} \)

Solution:
Use the small-angle approximation \( \arctan(x) \approx x \):
\[
\lim_{x \to 0} \frac{\arctan x + x^2}{2^{3x} - 3^{2x}} = \lim_{x \to 0} \frac{x + x^2}{2^{3x} - 3^{2x}}
\]

Use Taylor series for the exponentials:
\[
2^{3x} = e^{3x \ln 2} \approx 1 + 3x \ln 2, \quad 3^{2x} = e^{2x \ln 3} \approx 1 + 2x \ln 3
\]
\[
2^{3x} - 3^{2x} \approx (1 + 3x \ln 2) - (1 + 2x \ln 3) = x(3 \ln 2 - 2 \ln 3)
\]

Thus:
\[
\lim_{x \to 0} \frac{x + x^2}{x(3 \ln 2 - 2 \ln 3)} = \lim_{x \to 0} \frac{1 + x}{3 \ln 2 - 2 \ln 3} = \frac{1}{3 \ln 2 - 2 \ln 3}
\]

Answer:
\[
\boxed{\frac{1}{3 \ln 2 - 2 \ln 3}}
\]

---

#### r) \( \lim_{x \to \frac{\pi}{2}} \frac{\arctan(x - \pi/2)}{\pi - 2x} \)

Solution:
Let \( u = x - \frac{\pi}{2} \). As \( x \to \frac{\pi}{2} \), \( u \to 0 \). The limit becomes:
\[
\lim_{u \to 0} \frac{\arctan(u)}{\pi - 2\left(\frac{\pi}{2} + u\right)} = \lim_{u \to 0} \frac{\arctan(u)}{\pi - \pi - 2u} = \lim_{u \to 0} \frac{\arctan(u)}{-2u}
\]

Use the small-angle approximation \( \arctan(u) \approx u \):
\[
\lim_{u \to 0} \frac{u}{-2u} = -\frac{1}{2}
\]

Answer:
\[
\boxed{-\frac{1}{2}}
\]

---

#### s) \( \lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{2\arcsin x - \sin x} \)

Solution:
Use Taylor series expansions:
\[
e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \cdots, \quad e^{-2x} = 1 - 2x + \frac{(-2x)^2}{2!} - \cdots
\]
\[
e^{3x} - e^{-2x} = (1 + 3x + \frac{9x^2}{2} + \cdots) - (1 - 2x + \frac{4x^2}{2} - \cdots) = 5x + \frac{5x^2}{2} + \cdots
\]
\[
2\arcsin x - \sin x \approx 2x - x = x
\]
\[
\frac{e^{3x} - e^{-2x}}{2\arcsin x - \sin x} = \frac{5x + \frac{5x^2}{2} + \cdots}{x} = 5 + \frac{5x}{2} + \cdots
\]

As \( x \to 0 \):
\[
\lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{2\arcsin x - \sin x} = 5
\]

Answer:
\[
\boxed{5}
\]

---

#### t) \( \lim_{x \to 0} \frac{\ln(\cos 3x)}{\arctan 4x} \)

Solution:
Use the small-angle approximations \( \cos(3x) \approx 1 - \frac{(3x)^2}{2} \) and \( \arctan(4x) \approx 4x \):
\[
\ln(\cos 3x) \approx \ln\left(1 - \frac{9x^2}{2}\right) \approx -\frac{9x^2}{2}
\]
\[
\arctan(4x) \approx 4x
\]
\[
\frac{\ln(\cos 3x)}{\arctan 4x} \approx \frac{-\frac{9x^2}{2}}{4x} = -\frac{9x}{8}
\]

As \( x \to 0 \):
\[
\lim_{x \to 0} \frac{\ln(\cos 3x)}{\arctan 4x} = 0
\]

Answer:
\[
\boxed{0}
\]

---

#### u) \( \lim_{x \to 0} \frac{(1+x)^2 - (1+2x)}{x^2 + 4x^3} \)

Solution:
Expand the numerator:
\[
(1+x)^2 = 1 + 2x + x^2
\]
\[
(1+x)^2 - (1+2x) = (1 + 2x + x^2) - (1 + 2x) = x^2
\]

The limit becomes:
\[
\lim_{x \to 0} \frac{x^2}{x^2 + 4x^3} = \lim_{x \to 0} \frac{1}{1 + 4x} = 1
\]

Answer:
\[
\boxed{1}
\]

---

#### v) \( \lim_{x \to -1} \frac{(x^2 + 3x + 2)^2}{x^3 - 3x - 2} \)

Solution:
Factor the numerator and the denominator:
- Numerator: \( x^2 + 3x + 2 = (x+1)(x+2) \), so \( (x^2 + 3x + 2)^2 = ((x+1)(x+2))^2 \)
- Denominator: Test \( x = -1 \) as a root:
\[
(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0
\]
So, \( x+1 \) is a factor. Perform polynomial division or factorization to get:
\[
x^3 - 3x - 2 = (x+1)(x^2 - x - 2) = (x+1)(x-2)(x+1)
\]

Thus, the expression becomes:
\[
\lim_{x \to -1} \frac{((x+1)(x+2))^2}{(x+1)(x-2)(x+1)} = \lim_{x \to -1} \frac{(x+2)^2}{x-2}
\]

Substitute \( x = -1 \):
\[
\frac{(-1+2)^2}{-1-2} = \frac{1^2}{-3} = \frac{1}{-3} = -\frac{1}{3}
\]

Answer:
\[
\boxed{-\frac{1}{3}}
\]

---

#### w) \( \lim_{x \to 0} \frac{1 - \cos^4 x}{4x^2} \)

Solution:
Use the identity \( \cos^4 x = (\cos^2 x)^2 \) and the Pythagorean identity \( \cos^2 x = 1 - \sin^2 x \):
\[
1 - \cos^4 x = 1 - (1 - \sin^2 x)^2 = 1 - (1 - 2\sin^2 x + \sin^4 x) = 2\sin^2 x - \sin^4 x
\]

The limit becomes:
\[
\lim_{x \to 0} \frac{2\sin^2 x - \sin^4 x}{4x^2} = \lim_{x \to 0} \frac{\sin^2 x (2 - \sin^2 x)}{4x^2}
\]

Use the small-angle approximation \( \sin x \approx x \):
\[
\frac{\sin^2 x (2 - \sin^2 x)}{4x^2} \approx \frac{x^2 (2 - x^2)}{4x^2} = \frac{2 - x^2}{4}
\]

As \( x \to 0 \):
\[
\lim_{x \to 0} \frac{2 - x^2}{4} = \frac{2}{4} = \frac{1}{2}
\]

Answer:
\[
\boxed{\frac{1}{2}}
\]

---

#### x) \( \lim_{x \to 0} \frac{1 - \cos x}{2 \sin x} \)

Solution:
Use the small-angle approximations \( 1 - \cos x \approx \frac{x^2}{2} \) and \( \sin x \approx x \):
\[
\lim_{x \to 0} \frac{1 - \cos x}{2 \sin x} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{2x} = \lim_{x \to 0} \frac{x^2}{4x} = \lim_{x \to 0} \frac{x}{4} = 0
\]

Answer:
\[
\boxed{0}
\]

---

#### y) \( \lim_{x \to 0} \frac{x - \sin x}{e^x - e^{-x} - 2x} \)

Solution:
Use Taylor series expansions:
\[
x - \sin x \approx x - \left(x - \frac{x^3}{3!} + \cdots\right) = \frac{x^3}{6} + \cdots
\]
\[
e^x - e^{-x} \approx \left(1 + x + \frac{x^2}{2!} + \cdots\right) - \left(1 - x + \frac{x^2}{2!} - \cdots\right) = 2x + \frac{2x^3}{3!} + \cdots
\]
\[
e^x - e^{-x} - 2x \approx \frac{2x^3}{6} + \cdots = \frac{x^3}{3} + \cdots
\]

The limit becomes:
\[
\lim_{x \to 0} \frac{\frac{x^3}{6} + \cdots}{\frac{x^3}{3} + \cdots} = \lim_{x \to 0} \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}
\]

Answer:
\[
\boxed{\frac{1}{2}}
\]

---

#### z) \( \lim_{x \to 0} \frac{e^{3x} - 3x - 1}{\sin^2 x} \)

Solution:
Use Taylor series expansions:
\[
e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \cdots = 1 + 3x + \frac{9x^2}{2} + \cdots
\]
\[
e^{3x} - 3x - 1 \approx \frac{9x^2}{2} + \cdots
\]
\[
\sin^2 x \approx x^2
\]

The limit becomes:
\[
\lim_{x \to 0} \frac{\frac{9x^2}{2} + \cdots}{x^2} = \lim_{x \to 0} \frac{9x^2}{2x^2} = \frac{9}{2}
\]

Answer:
\[
\boxed{\frac{9}{2}}
\]

---

#### Z) \( \lim_{x \to 0} \frac{x^3}{x - \arctan x} \)

Solution:
Use the small-angle approximation \( \arctan x \approx x - \frac{x^3}{3} \):
\[
x - \arctan x \approx x - \left(x - \frac{x^3}{3}\right) = \frac{x^3}{3}
\]

The limit becomes:
\[
\lim_{x \to 0} \frac{x^3}{\frac{x^3}{3}} = \lim_{x \to 0} 3 = 3
\]

Answer:
\[
\boxed{3}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{lll}
\text{a)} & 0 & \text{j)} & 1 & \text{s)} & 5 \\
\text{b)} & \frac{18}{23} & \text{k)} & \frac{3}{2} & \text{t)} & 0 \\
\text{c)} & \frac{16}{9} & \text{l)} & 2 & \text{u)} & 1 \\
\text{d)} & \frac{\sqrt{2}}{8} & \text{m)} & \frac{\pi}{3} & \text{v)} & -\frac{1}{3} \\
\text{e)} & \ln(5) & \text{n)} & \frac{4}{\ln(3)} & \text{w)} & \frac{1}{2} \\
\text{f)} & \frac{6}{\ln(3)} & \text{o)} & 4 & \text{x)} & 0 \\
\text{g)} & 2 & \text{p)} & -1 & \text{y)} & \frac{1}{2} \\
\text{h)} & \infty & \text{q)} & \frac{1}{3 \ln 2 - 2 \ln 3} & \text{z)} & \frac{9}{2} \\
\text{i)} & -\frac{\sqrt{3}}{3} & \text{r)} & -\frac{1}{2} & \text{Z)} & 3 \\
\end{array}
}
\]