Properties of Limit Function

Understanding Properties of Limit Function

After learning the basic concept of limits, we will now delve into properties of limit functions that are very helpful in solving complex limit calculations. Imagine these properties as game rules that allow us to break complex limits into simpler parts.

These limit properties become an important foundation in calculus because they allow us to calculate limits without always having to use formal definitions or value tables. By understanding these properties, limit calculations become more efficient and systematic.

Basic Properties of Limits

Constant Property

The simplest property is the limit of a constant function. If $k$ is a constant, then:

This means, the limit of a constant is the constant itself. This makes sense because the value of a constant does not change with respect to the variable $x$.

Identity Property

For the identity function, the following holds:

When $x$ approaches $c$, the value of function $f(x) = x$ also approaches $c$.

Arithmetic Operation Properties

Suppose $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ where $L$ and $M$ are real numbers, then the following properties hold:

Addition and Subtraction Properties

The limit of the sum or difference of two functions equals the sum or difference of the limits of each function:

This property allows us to break complex limits into simpler parts.

Multiplication Property

The limit of the product of two functions equals the product of the limits of each function:

Multiplication by Constant Property

A constant can be factored out from the limit sign:

Division Property

The limit of the quotient of two functions equals the quotient of the limits of each function, provided the limit of the denominator is not zero:

with the condition $M \neq 0$.

Power and Root Properties

Power Property

The limit of a function raised to a power equals the power of the limit of the function:

where $n$ is a real number.

Root Property

The limit of the root of a function equals the root of the limit of the function:

Important conditions:

• If $n$ is odd: this property applies to all values of $L$
• If $n$ is even: $L \geq 0$ (cannot be negative because the even root of a negative number is not defined in real numbers)

Examples of Applying Limit Properties

Simple Example

Calculate $\lim_{x \to 2} (3x^2 + 5x - 1)$.

Solution:

Using limit properties:

Example with Fractions

Calculate $\lim_{x \to 4} \frac{x\sqrt{x}}{x^2 + 3}$.

Solution:

Using division and multiplication properties:

Now we substitute the value $x = 4$:

In decimal form: $\frac{8}{19} \approx 0.421$

Example with Roots

Calculate $\lim_{x \to 0} \sqrt{x^2 - 3x + 2}$.

Solution:

Using the root property (since $n = 2$ is even, we need to ensure the result inside the root is not negative):

Calculate the limit inside the root first:

Since $2 > 0$, we can use the root property:

Exercises

1. Calculate $\lim_{x \to 3} (2x^2 - 4x + 1)$

2. Calculate $\lim_{x \to 1} \frac{3x + 2}{x^2 + 1}$

3. Calculate $\lim_{x \to 4} \sqrt{x + 5}$

4. Calculate $\lim_{x \to 2} (x + 1)^3$

5. Calculate $\lim_{x \to 0} \frac{5x^2 + 3x}{2x + 1}$

Answer Key

1. Solution:

   Using addition and multiplication by constant properties:

   $$\lim_{x \to 3} (2x^2 - 4x + 1) = 2\lim_{x \to 3} x^2 - 4\lim_{x \to 3} x + \lim_{x \to 3} 1$$

   Substitute $x = 3$:

   $$= 2(3)^2 - 4(3) + 1 = 2(9) - 12 + 1 = 18 - 12 + 1 = 7$$

2. Solution:

   Using the division property:

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

   $$= \frac{3(1) + 2}{1^2 + 1} = \frac{5}{2}$$

   In decimal form: $\frac{5}{2} = 2.5$

3. Solution:

   Using the root property:

   $$\lim_{x \to 4} \sqrt{x + 5} = \sqrt{\lim_{x \to 4} (x + 5)}$$

   $$= \sqrt{4 + 5} = \sqrt{9} = 3$$

4. Solution:

   Using the power property:

   $$\lim_{x \to 2} (x + 1)^3 = \left[\lim_{x \to 2} (x + 1)\right]^3$$

   $$= (2 + 1)^3 = 3^3 = 27$$

5. Solution:

   Using the division property:

   $$\lim_{x \to 0} \frac{5x^2 + 3x}{2x + 1} = \frac{\lim_{x \to 0} (5x^2 + 3x)}{\lim_{x \to 0} (2x + 1)}$$

   $$= \frac{5(0)^2 + 3(0)}{2(0) + 1} = \frac{0}{1} = 0$$