Limit of Algebraic Function

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 $g(x) = \frac{x + 1}{x - 2}$.

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:

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

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:

• 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.

Factoring Technique

The most common method is to factor the numerator and denominator, then simplify.

Example: Calculate $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Direct substitution gives $\frac{0}{0}$. Let's factor:

Since $x$ approaches 2 (not equal to 2), we can cancel the factor $(x - 2)$:

Rationalization Technique

For limits involving radical forms, we often need to rationalize. Direct substitution yields an indeterminate form.

Example: Calculate $\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$

Direct substitution:

We rationalize by multiplying with the conjugate $\sqrt{x} + 1$. The purpose is to eliminate the radical form in the numerator using the formula $(a-b)(a+b) = a^2 - b^2$:

Since $x \neq 1$ (approaching 1), we can cancel $(x - 1)$:

Application of Limit Properties to Algebraic Functions

The limit properties we have learned can be applied systematically:

Combining Properties

Example: Calculate $\lim_{x \to 3} \frac{x^2 - 9}{x^2 + 2x - 15}$

Direct substitution:

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 $-3$.

Therefore:

Substitute $x = 3$: $\frac{3 + 3}{3 + 5} = \frac{6}{8} = \frac{3}{4}$

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.

Continuity and Limits

A function $f$ is said to be continuous at $x = c$ if:

1. $f(c)$ exists (is defined)
2. $\lim_{x \to c} f(x)$ exists
3. $\lim_{x \to c} f(x) = f(c)$

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 $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$

Answer Key

1. Solution:

   Since this is a polynomial function that is continuous at all points, we can substitute directly:

   $$\lim_{x \to 2} (x^4 - 5x^3 + x^2 - 7) = 2^4 - 5(2^3) + 2^2 - 7$$

   $$= 16 - 5(8) + 4 - 7 = 16 - 40 + 4 - 7 = -27$$

2. Solution:

   Since this is a rational function with a non-zero denominator at $x = 2$, use direct substitution:

   $$\lim_{x \to 2} \frac{x^2 - 6x + 8}{x^2 + 16x + 28} = \frac{2^2 - 6(2) + 8}{2^2 + 16(2) + 28}$$

   $$= \frac{4 - 12 + 8}{4 + 32 + 28} = \frac{0}{64} = 0$$

3. Solution:

   Direct substitution gives $\frac{0}{0}$. Let's factor the numerator:

   $$x^2 - 2x + 1 = (x - 1)^2$$

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

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 $\sqrt{x + 4} + 2$:

   $$\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$$

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

   $$= \lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2}$$

   $$= \frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$$