# Nakafa Learning Content
> For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.
URL: https://nakafa.com/en/subjects/mathematics/limit/limit-of-algebraic-function
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/limit/limit-of-algebraic-function/en.mdx
Calculate limits of polynomial and rational functions. Learn factoring, rationalization techniques, and solve indeterminate forms with worked examples.
---
## Understanding Limits of Algebraic Functions
Imagine you are driving a car towards a destination. The closer you get to your destination, the clearer you can see its details. In mathematics, **limits of algebraic functions** work in a similar way. Limits show the value approached by an algebraic function when its input variable approaches a certain value.
Algebraic functions are functions formed from combinations of algebraic operations such as addition, subtraction, multiplication, division, and exponentiation with rational exponents. Examples include polynomial functions like $$f(x) = x^2 + 3x - 2$$ and rational functions like $$g(x) = \frac{x + 1}{x - 2}$$.
Visible text: Algebraic functions are functions formed from combinations of algebraic operations such as addition, subtraction, multiplication, division, and exponentiation with rational exponents. Examples include polynomial functions like and rational functions like .
## Fundamental Properties of Algebraic Limits
To calculate limits of algebraic functions, we can use basic properties that are very helpful:
### Limits of Polynomial Functions
For polynomial functions that are **continuous** at all points, calculating limits is very simple. We can directly **substitute** the approaching value.
Let $$f(x) = x^4 - 5x^3 + x^2 - 7$$, then:
Visible text: Let , then:
```math
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^4 - 5x^3 + x^2 - 7)
```
Since polynomial functions are continuous at all points, we can substitute directly:
```math
= 2^4 - 5(2^3) + 2^2 - 7 = 16 - 40 + 4 - 7 = -27
```
### Limits of Rational Functions
Rational functions have the form $$\frac{P(x)}{Q(x)}$$ where $$P(x)$$ and $$Q(x)$$ are polynomials. Their limit calculation depends on the denominator value:
Visible text: Rational functions have the form where and are polynomials. Their limit calculation depends on the denominator value:
- **If the denominator is not zero:** Use direct substitution like polynomial functions.
- **If the denominator is zero:** We obtain an indeterminate form that requires algebraic manipulation.
## Handling Indeterminate Forms
When direct substitution yields the form $$\frac{0}{0}$$, we need to use special techniques.
Visible text: When direct substitution yields the form , we need to use special techniques.
### Factoring Technique
The most common method is to factor the numerator and denominator, then simplify.
**Example:** Calculate $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$
Visible text: **Example:** Calculate
Direct substitution gives $$\frac{0}{0}$$. Let's factor:
Visible text: Direct substitution gives . Let's factor:
Component: MathContainer
Children:
```math
x^2 - 4 = (x + 2)(x - 2)
```
```math
\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x + 2)(x - 2)}{x - 2}
```
Since $$x$$ approaches $$2$$ (not equal to $$2$$), we can cancel the factor $$(x - 2)$$:
Visible text: Since approaches (not equal to ), we can cancel the factor :
```math
= \lim_{x \to 2} (x + 2) = 2 + 2 = 4
```
### Rationalization Technique
For limits involving radical forms, we often need to rationalize. Direct substitution yields an indeterminate form.
**Example:** Calculate $$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$$
Visible text: **Example:** Calculate
Direct substitution:
Component: MathContainer
Children:
```math
\frac{\sqrt{1} - 1}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0} \text{ (indeterminate form)}
```
We rationalize by multiplying with the **conjugate** $$\sqrt{x} + 1$$. The purpose is to eliminate the radical form in the numerator using the formula $$(a-b)(a+b) = a^2 - b^2$$:
Visible text: We rationalize by multiplying with the **conjugate** . The purpose is to eliminate the radical form in the numerator using the formula :
Component: MathContainer
Children:
```math
\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1}
```
```math
= \lim_{x \to 1} \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)}
```
```math
= \lim_{x \to 1} \frac{(\sqrt{x})^2 - 1^2}{(x - 1)(\sqrt{x} + 1)} = \lim_{x \to 1} \frac{x - 1}{(x - 1)(\sqrt{x} + 1)}
```
Since $$x \neq 1$$ (approaching $$1$$), we can cancel $$(x - 1)$$:
Visible text: Since (approaching ), we can cancel :
```math
= \lim_{x \to 1} \frac{1}{\sqrt{x} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}
```
## Application of Limit Properties to Algebraic Functions
The limit properties we have learned can be applied systematically:
### Combining Properties
**Example:** Calculate $$\lim_{x \to 3} \frac{x^2 - 9}{x^2 + 2x - 15}$$
Visible text: **Example:** Calculate
Direct substitution:
Component: MathContainer
Children:
```math
\frac{3^2 - 9}{3^2 + 2(3) - 15} = \frac{9 - 9}{9 + 6 - 15} = \frac{0}{0} \text{ (indeterminate form)}
```
Let's factor both. For $$x^2 + 2x - 15$$, we find two numbers that when multiplied give $$-15$$ and when added give $$2$$. Those numbers are $$5$$ and $$-3$$.
Visible text: Let's factor both. For , we find two numbers that when multiplied give and when added give . Those numbers are and .
Component: MathContainer
Children:
```math
x^2 - 9 = (x - 3)(x + 3)
```
```math
x^2 + 2x - 15 = x^2 + 5x - 3x - 15 = x(x + 5) - 3(x + 5) = (x - 3)(x + 5)
```
Therefore:
Component: MathContainer
Children:
```math
\lim_{x \to 3} \frac{(x - 3)(x + 3)}{(x - 3)(x + 5)} = \lim_{x \to 3} \frac{x + 3}{x + 5}
```
Substitute $$x = 3$$: $$\frac{3 + 3}{3 + 5} = \frac{6}{8} = \frac{3}{4}$$
Visible text: Substitute :
> If the numerator is non-zero and the denominator is zero (like $$\frac{a}{0}$$ with $$a \neq 0$$), the limit approaches infinity. If both numerator and denominator are zero (the form $$\frac{0}{0}$$), use factoring or rationalization techniques.
Visible text: > If the numerator is non-zero and the denominator is zero (like with ), the limit approaches infinity. If both numerator and denominator are zero (the form ), use factoring or rationalization techniques.
## Continuity and Limits
A function $$f$$ is said to be **continuous** at $$x = c$$ if:
Visible text: A function is said to be **continuous** at if:
1. $$f(c)$$ exists (is defined)
2. $$\lim_{x \to c} f(x)$$ exists
3. $$\lim_{x \to c} f(x) = f(c)$$
Visible text: 1. exists (is defined)
2. exists
3.
Polynomial functions are continuous at all points, while rational functions are continuous at all points except where their denominator is zero.
## Exercises
1. Calculate $$\lim_{x \to 2} (x^4 - 5x^3 + x^2 - 7)$$
2. Calculate $$\lim_{x \to 2} \frac{x^2 - 6x + 8}{x^2 + 16x + 28}$$
3. Calculate $$\lim_{x \to 1} \frac{x^2 - 2x + 1}{x - 1}$$
4. Determine whether the function $$f(x) = x^2 - 2x + 1$$ is continuous at $$x = 1$$
5. Calculate $$\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$$
Visible text: 1. Calculate
2. Calculate
3. Calculate
4. Determine whether the function is continuous at
5. Calculate
### Answer Key
1. **Solution:**
Since this is a polynomial function that is continuous at all points, we can substitute directly:
```math
\lim_{x \to 2} (x^4 - 5x^3 + x^2 - 7) = 2^4 - 5(2^3) + 2^2 - 7
```
```math
= 16 - 5(8) + 4 - 7 = 16 - 40 + 4 - 7 = -27
```
2. **Solution:**
Since this is a rational function with a non-zero denominator at $$x = 2$$, use direct substitution:
```math
\lim_{x \to 2} \frac{x^2 - 6x + 8}{x^2 + 16x + 28} = \frac{2^2 - 6(2) + 8}{2^2 + 16(2) + 28}
```
```math
= \frac{4 - 12 + 8}{4 + 32 + 28} = \frac{0}{64} = 0
```
3. **Solution:**
Direct substitution gives $$\frac{0}{0}$$. Let's factor the numerator:
```math
x^2 - 2x + 1 = (x - 1)^2
```
```math
\lim_{x \to 1} \frac{(x - 1)^2}{x - 1} = \lim_{x \to 1} (x - 1) = 1 - 1 = 0
```
4. **Solution:**
To check continuity at $$x = 1$$, check three conditions:
- **Condition $$1$$ (function is defined):** $$f(1) = 1^2 - 2(1) + 1 = 1 - 2 + 1 = 0$$ ✓ (exists)
- **Condition $$2$$ (limit exists):** Since it's a polynomial function, $$\lim_{x \to 1} f(x) = f(1) = 0$$ ✓ (exists)
- **Condition $$3$$ (limit equals function value):** $$\lim_{x \to 1} f(x) = f(1) = 0$$ ✓ (equal)
Since all three continuity conditions are satisfied, the function is **continuous** at $$x = 1$$.
5. **Solution:**
Direct substitution: $$\frac{\sqrt{0 + 4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}$$ (indeterminate form).
Use rationalization by multiplying with the conjugate $$\sqrt{x + 4} + 2$$:
```math
\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}
```
```math
= \lim_{x \to 0} \frac{(\sqrt{x + 4})^2 - 2^2}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)}
```
```math
= \lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}
```
```math
= \frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
```
Visible text: 1. **Solution:**
Since this is a polynomial function that is continuous at all points, we can substitute directly:
2. **Solution:**
Since this is a rational function with a non-zero denominator at , use direct substitution:
3. **Solution:**
Direct substitution gives . Let's factor the numerator:
4. **Solution:**
To check continuity at , check three conditions:
- **Condition (function is defined):** ✓ (exists)
- **Condition (limit exists):** Since it's a polynomial function, ✓ (exists)
- **Condition (limit equals function value):** ✓ (equal)
Since all three continuity conditions are satisfied, the function is **continuous** at .
5. **Solution:**
Direct substitution: (indeterminate form).
Use rationalization by multiplying with the conjugate :