

# Limits as a fundamental concept

Limits are a fundamental concept in calculus. They are used to describe the behavior of a function as its input approaches a certain value. A limit of a function $f(x)$ as $x$ approaches a value $a$ is denoted as $\lim_{x \to a} f(x)$.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$. To find the limit as $x$ approaches $2$, we can plug in $2$ into the function and evaluate:

$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \frac{(2)^2 - 4}{2 - 2} = \frac{0}{0}$$

The limit is undefined because the function is not defined at $x = 2$.

## Exercise
Find the limit of the function $f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4}$ as $x$ approaches $2$.


Solve the limit by plugging in $2$ into the function and simplifying.

### Solution
$$\lim_{x \to 2} \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4} = \frac{(2)^3 - 6(2)^2 + 11(2) - 6}{(2)^2 - 4} = \frac{0}{0}$$

The limit is undefined because the function is not defined at $x = 2$.

# Algebraic limits and limits at infinity

Algebraic limits are limits that can be computed using algebraic manipulation. These limits are often found when dealing with polynomials and rational functions.

Consider the function $f(x) = \frac{x^2 + 2x + 1}{x + 1}$. To find the limit as $x$ approaches $-1$, we can plug in $-1$ into the function and evaluate:

$$\lim_{x \to -1} \frac{x^2 + 2x + 1}{x + 1} = \frac{(-1)^2 + 2(-1) + 1}{-1 + 1} = \frac{0}{0}$$

The limit is undefined because the function is not defined at $x = -1$.

## Exercise
Instructions:

Find the limit of the function $f(x) = \frac{x^3 - 3x^2 + 3x + 1}{x^2 - 1}$ as $x$ approaches $1$.


Solve the limit by plugging in $1$ into the function and simplifying.

### Solution
$$\lim_{x \to 1} \frac{x^3 - 3x^2 + 3x + 1}{x^2 - 1} = \frac{(1)^3 - 3(1)^2 + 3(1) + 1}{(1)^2 - 1} = \frac{1}{0}$$

The limit is undefined because the function is not defined at $x = 1$.

# Limit laws and their applications

Limit laws are rules that allow us to manipulate limits algebraically. These laws are useful for solving limits of more complex functions.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$ and the function $g(x) = x + 2$. The limit of the sum of these functions as $x$ approaches $2$ can be found using the limit laws:

$$\lim_{x \to 2} (f(x) + g(x)) = \lim_{x \to 2} \frac{x^2 - 4}{x - 2} + \lim_{x \to 2} (x + 2)$$

$$= \lim_{x \to 2} \frac{x^2 - 4}{x - 2} + \lim_{x \to 2} x + \lim_{x \to 2} 2$$

$$= \lim_{x \to 2} \frac{x^2 - 4}{x - 2} + 2$$

$$= \frac{(2)^2 - 4}{2 - 2} + 2 = \frac{0}{0} + 2 = \frac{2}{0}$$

The limit is undefined because the function is not defined at $x = 2$.

## Exercise
Instructions:

Find the limit of the sum of the functions $f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4}$ and $g(x) = x^2 - 4$ as $x$ approaches $2$.


Solve the limit by using the limit laws and simplifying.

### Solution
$$\lim_{x \to 2} (f(x) + g(x)) = \lim_{x \to 2} \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4} + \lim_{x \to 2} (x^2 - 4)$$

$$= \lim_{x \to 2} \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4} + \lim_{x \to 2} x^2 - \lim_{x \to 2} 4$$

$$= \lim_{x \to 2} \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4} + (2)^2 - 4$$

$$= \frac{(2)^3 - 6(2)^2 + 11(2) - 6}{(2)^2 - 4} + (2)^2 - 4$$

$$= \frac{0}{0} + 4 = \frac{4}{0}$$

The limit is undefined because the function is not defined at $x = 2$.

# Limit approximations and infinite series

Limit approximations are used to find the limit of a function by approximating it with simpler functions. Infinite series are a type of limit approximation that involve the sum of a sequence of terms.

Consider the function $f(x) = \frac{1}{x - 1}$. To find the limit as $x$ approaches $1$, we can use a limit approximation with a geometric series:

$$\lim_{x \to 1} \frac{1}{x - 1} = \lim_{x \to 1} \frac{1}{1 - (1 - x)} = \lim_{x \to 1} \frac{1}{1 - (1 - x)} = \lim_{x \to 1} \sum_{n = 0}^{\infty} (1 - x)^n$$

$$= \lim_{x \to 1} \sum_{n = 0}^{\infty} \left(\frac{1 - x}{1}\right)^n = \lim_{x \to 1} \sum_{n = 0}^{\infty} (1 - x)^n$$

The limit is undefined because the function is not defined at $x = 1$.

## Exercise
Instructions:

Find the limit of the function $f(x) = \frac{x^2 - 4}{x - 2}$ as $x$ approaches $2$ using a limit approximation with a geometric series.


Solve the limit by using a geometric series and simplifying.

### Solution
$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{1}{1 - (1 - x)} = \lim_{x \to 2} \frac{1}{1 - (1 - x)} = \lim_{x \to 2} \sum_{n = 0}^{\infty} (1 - x)^n$$

$$= \lim_{x \to 2} \sum_{n = 0}^{\infty} \left(\frac{1 - x}{1}\right)^n = \lim_{x \to 2} \sum_{n = 0}^{\infty} (1 - x)^n$$

The limit is undefined because the function is not defined at $x = 2$.

# Epsilon-delta definition of a limit

The epsilon-delta definition of a limit is a rigorous way to define the limit of a function. It states that the limit of a function $f(x)$ as $x$ approaches $a$ is $L$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$. To show that the limit as $x$ approaches $2$ is undefined, we can use the epsilon-delta definition:

Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^2 - 4 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

## Exercise
Instructions:

Show that the limit of the function $f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4}$ as $x$ approaches $2$ is undefined using the epsilon-delta definition.


Solve the problem by using the epsilon-delta definition and simplifying.

### Solution
Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^3 - 6(2)^2 + 11(2) - 6 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

# Working with epsilon-delta proofs

Epsilon-delta proofs are used to prove theorems and properties in calculus. They involve choosing appropriate values for $\epsilon$ and $\delta$ to satisfy the epsilon-delta definition.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$. To show that the limit as $x$ approaches $2$ is undefined, we can use the epsilon-delta definition:

Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^2 - 4 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

## Exercise
Instructions:

Prove that the limit of the function $f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4}$ as $x$ approaches $2$ is undefined using the epsilon-delta definition.


Solve the problem by using the epsilon-delta definition and simplifying.

### Solution
Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^3 - 6(2)^2 + 11(2) - 6 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

# Continuity and its properties

Continuity is a property of a function that states that the function is defined at a given point and that the limit of the function as $x$ approaches that point exists.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$. To show that the function is not continuous at $x = 2$, we can use the epsilon-delta definition:

Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^2 - 4 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

## Exercise
Instructions:

Show that the function $f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4}$ is not continuous at $x = 2$ using the epsilon-delta definition.


Solve the problem by using the epsilon-delta definition and simplifying.

### Solution
Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^3 - 6(2)^2 + 11(2) - 6 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

# Discontinuities and removable, vertical, and jump discontinuities

Discontinuities are points in a function where the limit does not exist or is undefined. They can be classified into removable, vertical, and jump discontinuities.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$. To find the discontinuity of the function, we can use the epsilon-delta definition:

Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^2 - 4 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

## Exercise
Instructions:

Find the discontinuity of the function $f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 4}$ using the epsilon-delta definition.


Solve the problem by using the epsilon-delta definition and simplifying.

### Solution
Let $L$ be the limit of $f(x)$ as $x$ approaches $2$. For every $\epsilon > 0$, let $\delta > 0$ be given.

Choose $x = 2$. Then $|f(x) - L| = |(2)^3 - 6(2)^2 + 11(2) - 6 - L| = |-2 - L| = |2 + L|$.

If we choose $\epsilon = 2 + L$, then $|f(x) - L| = \epsilon$.

Since we can always find an $\epsilon$ that satisfies the epsilon-delta definition, the limit of $f(x)$ as $x$ approaches $2$ is undefined.

# The intermediate value theorem

The intermediate value theorem is a theorem in calculus that states that if a function is continuous on an interval $[a, b]$ and takes on different values at the endpoints, then there exists at least one value $c$ in the interval $(a, b)$ such that $f(c) = 0$.

Consider the function $f(x) = x^2 - 4$. To find a value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$, we can use the intermediate value theorem:

Since $f(-2) = (-2)^2 - 4 = -8$ and $f(2) = (2)^2 - 4 = 0$, there exists at least one value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$.

## Exercise
Instructions:

Find a value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$ using the intermediate value theorem.


Solve the problem by using the intermediate value theorem and simplifying.

### Solution
Since $f(-2) = (-2)^2 - 4 = -8$ and $f(2) = (2)^2 - 4 = 0$, there exists at least one value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$.

# Applications of the intermediate value theorem

The intermediate value theorem has many applications in calculus. It can be used to solve equations, find roots of functions, and determine the existence of solutions to differential equations.

Consider the function $f(x) = x^2 - 4$. To find a value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$, we can use the intermediate value theorem:

Since $f(-2) = (-2)^2 - 4 = -8$ and $f(2) = (2)^2 - 4 = 0$, there exists at least one value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$.

## Exercise
Instructions:

Find a value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$ using the intermediate value theorem.


Solve the problem by using the intermediate value theorem and simplifying.

### Solution
Since $f(-2) = (-2)^2 - 4 = -8$ and $f(2) = (2)^2 - 4 = 0$, there exists at least one value $c$ in the interval $(-2, 2)$ such that $f(c) = 0$.

# Derivatives and their properties

Derivatives are used to find the rate of change of a function with respect to its input. They have many properties that can be used to analyze and solve problems in calculus.

Consider the function $f(x) = x^2 - 4$. To find the derivative of the function, we can use the power rule:

$$f'(x) = \frac{d}{dx} (x^2 - 4) = 2x$$

## Exercise
Instructions:

Find the derivative of the function $f(x) = x^3 - 6x^2 + 11x - 6$ using the power rule.


Solve the problem by using the power rule and simplifying.

### Solution
$$f'(x) = \frac{d}{dx} (x^3 - 6x^2 + 11x - 6) = 3x^2 - 12x + 11$$