Limit of Algebraic Function

Choosing a Method from the Algebraic Expression

For an algebraic limit, the form of the expression determines which operation is justified. A polynomial can be evaluated by direct substitution. A rational or radical expression may instead contain a removable factor, a domain boundary, or a vertical asymptote that must be analyzed first.

Algebraic functions are built from operations such as addition, subtraction, multiplication, division, and powers with rational exponents. Examples include the polynomial $f(x) = x^2 + 3x - 2$ and the rational function $g(x) = \frac{x + 1}{x - 2}$. We will use substitution as a diagnostic step, then factor, rationalize, or check one-sided behavior when substitution does not settle the limit.

Properties of Algebraic Limits

An algebraic limit can be reduced to limits of simpler terms. Applying the same rules repeatedly means working term by term, and each step handles one simple part of the whole expression.

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. Direct substitution tells us which method to try next:

• If the denominator is not zero: Use direct substitution like polynomial functions.

• If the denominator is zero: Check the numerator, signs, and one-sided behavior. Only $\frac{0}{0}$ is an indeterminate form that may be resolved by algebraic simplification. A nonzero numerator over zero is not indeterminate.

Handling Indeterminate Forms

When direct substitution yields the form $\frac{0}{0}$, we need to use special techniques. Factoring and rationalizing both change the expression until the common factor that produces the zero in the denominator cancels, and then substitution works again.

Factoring Technique

Factor the numerator and denominator, then cancel a common factor when it is nonzero near the limit point.

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

Direct substitution gives $\frac{0}{0}$. 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 rules can be applied one after another until direct substitution produces a defined value. Each step uses one rule, and the expression becomes simpler at every stage.

Factoring the Numerator and Denominator

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

Direct substitution:

Direct substitution gives the indeterminate form $\frac00$: both numerator and denominator are zero.

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 $-3$.

The calculation gives:

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

If direct substitution gives $\frac{a}{0}$ with $a \neq 0$, that result alone is not enough. Check the sign and both one-sided limits. For example:

Because the one-sided limits differ, $\lim_{x \to 2}\frac{1}{x-2}$ does not exist as a two-sided limit. By contrast, if substitution gives $\frac{0}{0}$, first factor, rationalize, or otherwise simplify the expression.

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 every real number. A rational function is continuous at each point in its domain, namely wherever its denominator is nonzero.

Exercises

Each problem asks for a limit of an algebraic expression. Identify the form first, because a zero over zero result needs factoring or rationalising.

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

Worked Solutions

1. Solution:

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

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

   $$\begin{aligned} &= 16 - 5(8) + 4 - 7 \\ &= 16 - 40 + 4 - 7 \\ &= -27 \end{aligned}$$

2. Solution:

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

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

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

3. Solution:

   Direct substitution gives $\frac{0}{0}$. Factor the numerator:

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

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

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

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

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

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

   $$\begin{aligned} &= \frac{1}{\sqrt{0 + 4} + 2} \\ &= \frac{1}{\sqrt{4} + 2} \\ &= \frac{1}{2 + 2} \\ &= \frac{1}{4} \end{aligned}$$