# Nakafa Framework: LLM

URL: /en/subject/high-school/12/mathematics/limit/concept-of-limit-function
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/12/mathematics/limit/concept-of-limit-function/en.mdx

Output docs content for large language models.

---

export const metadata = {
    title: "Concept of Limit Function",
    description: "Understand function limits through intuitive examples and value tables. Learn left/right limits, formal definitions, and solve indeterminate forms.",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "05/26/2025",
    subject: "Limits",
};

## Understanding Limits Intuitively

Imagine you are walking towards your house door. The closer you get to the door, the clearer you can see the details of the door. In mathematics, **limits** work in a similar way. Limits describe the value that a function approaches when its input variable approaches a certain value.

The concept of limits is very fundamental in calculus because it becomes the foundation for understanding derivatives, integrals, and function continuity. Limits help us understand function behavior around certain points, even when the function is not defined exactly at that point.

## Approach Through Value Tables

To understand limits more concretely, let's see how function values change when the variable approaches a certain point. Suppose we have function <InlineMath math="f(x) = x + 2" /> and want to see what happens when <InlineMath math="x" /> approaches 3.

| <InlineMath math="x" /> | <InlineMath math="2.9" /> | <InlineMath math="2.99" /> | <InlineMath math="2.999" /> | ... | <InlineMath math="3.001" /> | <InlineMath math="3.01" /> | <InlineMath math="3.1" /> |
|-------------|-----|------|-------|-----|-------|------|-----|
| <InlineMath math="f(x)" /> | <InlineMath math="4.9" /> | <InlineMath math="4.99" /> | <InlineMath math="4.999" /> | ... | <InlineMath math="5.001" /> | <InlineMath math="5.01" /> | <InlineMath math="5.1" /> |

From the table above, we can see that when <InlineMath math="x" /> approaches 3 from the left (values <InlineMath math="x < 3" />) and from the right (values <InlineMath math="x > 3" />), the value of <InlineMath math="f(x)" /> approaches 5. **This approaching value is called the limit**.

## Formal Definition of Limits

Mathematically, limits can be defined as follows:

<BlockMath math="\lim_{x \to c} f(x) = L" />

This definition is read as "**the limit of <InlineMath math="f(x)" /> as <InlineMath math="x" /> approaches <InlineMath math="c" /> equals <InlineMath math="L" />**".

The conditions for this limit to exist are:
- **Left limit** and **right limit** must exist
- Left limit must be **equal to** the right limit
- The limit value is <InlineMath math="L" />

More formally, left and right limits can be written as:

<div className="flex flex-col gap-4">
<BlockMath math="\lim_{x \to c^-} f(x) = L \quad \text{(left limit: } x \text{ approaches } c \text{ from the left)}" />
<BlockMath math="\lim_{x \to c^+} f(x) = L \quad \text{(right limit: } x \text{ approaches } c \text{ from the right)}" />
</div>

If both limits are equal, then <InlineMath math="\lim_{x \to c} f(x) = L" />. If they are different, then the limit does not exist.

## Application of Limits

### Simple Example

Find <InlineMath math="\lim_{x \to 4} (2x - 1)" />.

**Solution:**

Since the function <InlineMath math="f(x) = 2x - 1" /> is continuous at <InlineMath math="x = 4" />, we can directly substitute:

<BlockMath math="\lim_{x \to 4} (2x - 1) = 2(4) - 1 = 8 - 1 = 7" />

### Example with Indeterminate Form

Find <InlineMath math="\lim_{x \to 2} \frac{x^2 - 4}{x - 2}" />.

**Solution:**

If we substitute <InlineMath math="x = 2" /> directly, we get the indeterminate form <InlineMath math="\frac{0}{0}" />. We need to simplify first by factoring:

<div className="flex flex-col gap-4">
<BlockMath math="x^2 - 4 = (x + 2)(x - 2)" />
<BlockMath math="\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x + 2)(x - 2)}{x - 2}" />
</div>

Since we are calculating the limit when <InlineMath math="x" /> **approaches** 2 (not **equals** 2), then <InlineMath math="x \neq 2" /> and we can cancel <InlineMath math="(x - 2)" />:

<BlockMath math="= \lim_{x \to 2} (x + 2) = 2 + 2 = 4" />

## Basic Properties of Limits

Some important properties that facilitate limit calculations:

1. **Linearity Property:**

    <InlineMath math="\lim_{x \to c} [af(x) + bg(x)] = a\lim_{x \to c} f(x) + b\lim_{x \to c} g(x)" />

2. **Multiplication Property:**

    <InlineMath math="\lim_{x \to c} [f(x) \cdot g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)" />

3. **Division Property:**

    <InlineMath math="\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}" /> provided <InlineMath math="\lim_{x \to c} g(x) \neq 0" />

## Exercises

1. Find <InlineMath math="\lim_{x \to 3} (x^2 + 2x - 1)" />

2. Find <InlineMath math="\lim_{x \to 1} \frac{x^2 - 1}{x - 1}" />

3. Find <InlineMath math="\lim_{x \to 0} \frac{\sin x}{x}" /> (use trigonometric limit theorem)

4. If <InlineMath math="f(x) = \begin{cases} x + 1, & x < 2 \\ 3x - 2, & x \geq 2 \end{cases}" />, find <InlineMath math="\lim_{x \to 2} f(x)" />

### Answer Key

1. **Solution:**

   Since polynomial functions are continuous at all points, we can substitute directly:

   <BlockMath math="\lim_{x \to 3} (x^2 + 2x - 1) = 3^2 + 2(3) - 1 = 9 + 6 - 1 = 14" />

2. **Solution:**

   Direct substitution yields the form <InlineMath math="\frac{0}{0}" />. We factor first:

   <div className="flex flex-col gap-4">
   <BlockMath math="x^2 - 1 = (x + 1)(x - 1)" />
   <BlockMath math="\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x + 1)(x - 1)}{x - 1}" />
   </div>

   Since <InlineMath math="x" /> approaches 1 (not equal to 1), then <InlineMath math="x \neq 1" /> and we can cancel <InlineMath math="(x - 1)" />:

   <BlockMath math="= \lim_{x \to 1} (x + 1) = 1 + 1 = 2" />

3. **Solution:**

   This is a **fundamental trigonometric limit** that is very important in calculus. This limit cannot be solved by direct substitution because it would yield the form <InlineMath math="\frac{0}{0}" />. However, based on the proven trigonometric limit theorem:

   <BlockMath math="\lim_{x \to 0} \frac{\sin x}{x} = 1" />

   **Note:** <InlineMath math="x" /> is in radians, not degrees.

4. **Solution:**

   For piecewise functions (defined with different rules), we must check left and right limits separately:

   **Left limit** (when <InlineMath math="x" /> approaches 2 from the left, so <InlineMath math="x < 2" />):

   <BlockMath math="\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 2 + 1 = 3" />

   **Right limit** (when <InlineMath math="x" /> approaches 2 from the right, so <InlineMath math="x \geq 2" />):
   
   <BlockMath math="\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 2) = 3(2) - 2 = 6 - 2 = 4" />
   
   Since <InlineMath math="\lim_{x \to 2^-} f(x) = 3 \neq 4 = \lim_{x \to 2^+} f(x)" />, then <InlineMath math="\lim_{x \to 2} f(x)" /> **does not exist**.