Properties of Function Limits

Sahabat Smaridasa - Mathematical Concepts

To make it easier to determine the limit value of a function, we need what are called properties of function limits. These properties are theorems used to solve limits of functions. There are various methods to compute limits, one of which is direct substitution, which we will use in this article. You may also read the material on "Definition of Function Limits".

Evaluating Limits by Direct Substitution

The substitution method means directly substituting the value \(x\) into the function \(f(x)\). For example: \(\displaystyle \lim_{x \to a} f(x) = f(a)\).

📘 Example 1 :

Determine the limit values of the following forms:

a) \(\displaystyle \lim_{x \to 2} (2x + 1)\)     b) \(\displaystyle \lim_{x \to -1} (2x - 1)\)

Solution:

a) \(\displaystyle \lim_{x \to 2} (2x + 1) = 2(2) + 1 = 4 + 1 = 5\).
Hence, \(\displaystyle \lim_{x \to 2} (2x + 1) = 5\).

b) \(\displaystyle \lim_{x \to -1} (2x - 1) = 2(-1) - 1 = -2 - 1 = -3\).
Hence, \(\displaystyle \lim_{x \to -1} (2x - 1) = -3\).

Properties of Function Limits

The following are the properties of function limits (assuming all individual limits exist):

• i) \(\displaystyle \lim_{x \to a} k = k\), where \(k\) is a constant.
• ii) \(\displaystyle \lim_{x \to a} [k \cdot f(x)] = k \cdot \lim_{x \to a} f(x)\).
• iii) \(\displaystyle \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)\).
• iv) \(\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = \left( \lim_{x \to a} f(x) \right) \left( \lim_{x \to a} g(x) \right)\).
• v) \(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{ \lim\limits_{x \to a} f(x) }{ \lim\limits_{x \to a} g(x) }\), provided \(\lim\limits_{x \to a} g(x) \neq 0\).
• vi) \(\displaystyle \lim_{x \to a} [f(x)]^n = \left[ \lim_{x \to a} f(x) \right]^n\) for any integer \(n\).
• vii) \(\displaystyle \lim_{x \to a} \sqrt{f(x)} = \sqrt{ \lim_{x \to a} f(x) }\), provided the limit inside the root is non-negative (for even roots).

📘 Example 2 :

Determine the following limits using the properties of limits:

a) \(\displaystyle \lim_{x \to 2} 5\)    b) \(\displaystyle \lim_{x \to 3} 2x^3\)    c) \(\displaystyle \lim_{x \to 1} (x^2 + x)\)    d) \(\displaystyle \lim_{x \to -1} (x^2 - 3x)\)

e) \(\displaystyle \lim_{x \to -2} (x^3 \cdot x^2)\)    f) \(\displaystyle \lim_{x \to 3} \frac{x+1}{x^2-1}\)    g) \(\displaystyle \lim_{x \to 2} (2x^2 + 3)^9\)    h) \(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1}\)

Solution (using limit properties):

a) \(\displaystyle \lim_{x \to 2} 5 = 5\).

b) \(\displaystyle \lim_{x \to 3} 2x^3 = 2 \cdot \lim_{x \to 3} x^3 = 2 \cdot 3^3 = 2 \cdot 27 = 54\).

c) \(\displaystyle \lim_{x \to 1} (x^2 + x) = \lim_{x \to 1} x^2 + \lim_{x \to 1} x = 1^2 + 1 = 1 + 1 = 2\).

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

e) \(\displaystyle \lim_{x \to -2} (x^3 \cdot x^2) = \left( \lim_{x \to -2} x^3 \right) \left( \lim_{x \to -2} x^2 \right) = (-2)^3 \cdot (-2)^2 = (-8) \cdot 4 = -32\).

f) \(\displaystyle \lim_{x \to 3} \frac{x+1}{x^2-1} = \frac{ \lim_{x \to 3} (x+1) }{ \lim_{x \to 3} (x^2-1) } = \frac{3+1}{3^2 - 1} = \frac{4}{9-1} = \frac{4}{8} = \frac{1}{2}\).

g) \(\displaystyle \lim_{x \to 2} (2x^2 + 3)^9 = \left( \lim_{x \to 2} (2x^2 + 3) \right)^9 = \left( 2 \cdot 2^2 + 3 \right)^9 = (2 \cdot 4 + 3)^9 = (8+3)^9 = 11^9\).

h) \(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1} = \sqrt{ \lim_{x \to 3} (x^2 - 1) } = \sqrt{3^2 - 1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}\).

💡 Note: All solutions above use the limit properties listed. Substitution works because these functions are continuous at the given points.

📘 Example 3 :

Evaluate \(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1}\) using properties (alternate illustration).

Solution:

\(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1} = \sqrt{ \lim_{x \to 3} (x^2 - 1) } = \sqrt{3^2 - 1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}\).

Thus, the limit equals \(2\sqrt{2}\).