Evaluating Limits Algebraically

You have probably noticed that making tables of values or staring at graphs can feel tedious when trying to find a limit. The good news is that mathematicians are, at heart, lazy in the best possible way: they developed algebraic techniques precisely so they would not have to plug in endless numbers. Once you learn these methods, you will be able to find most limits with just a few lines of algebra. It feels a bit like learning a magic trick, except the trick actually works every time and you understand exactly why.

The core idea is simple: if a function behaves nicely at a point, just plug in the number and you are done. When things get messy (read: when plugging in gives you something nonsensical like $\frac{0}{0}$), you have a toolkit of algebraic moves to clean up the expression first. That is the entire game plan for this lesson.

Core Concepts

Direct Substitution: The Easy Case

The simplest way to evaluate a limit is to just substitute the value of $x$ directly into the function. If you get a real number, congratulations, you have found your limit.

$$\lim_{x \to a} f(x) = f(a)$$

This works whenever the function is continuous at the point $a$. For now, think of continuous as meaning “the function has no holes, jumps, or vertical asymptotes at that point.” Most functions you encounter, including polynomials, are continuous everywhere.

For example, to find $\lim_{x \to 3} (x^2 + 2x - 1)$, you simply substitute $x = 3$:

$$\lim_{x \to 3} (x^2 + 2x - 1) = 3^2 + 2(3) - 1 = 9 + 6 - 1 = 14$$

That is it. No tricks, no complications. When direct substitution works, use it and move on.

Limit Laws: Breaking Problems Apart

Just as you can break apart arithmetic expressions (adding numbers one at a time, for instance), you can break apart limits using a set of reliable rules called limit laws. These laws assume that the individual limits exist.

Sum Law: $\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$

Difference Law: $\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$

Constant Multiple Law: $\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x)$

Product Law: $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$

Quotient Law: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$, provided $\lim_{x \to a} g(x) \neq 0$

Power Law: $\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a} f(x)\right]^n$

These laws let you handle complicated expressions piece by piece. Want to find $\lim_{x \to 2} (3x^2 - 5x)$? You can split it into $3 \cdot \lim_{x \to 2} x^2 - 5 \cdot \lim_{x \to 2} x$, evaluate each piece, and combine them.

Indeterminate Forms: When Direct Substitution Fails

Sometimes, substituting directly gives you a result that does not make sense as an answer. The most common troublemaker is the form $\frac{0}{0}$.

If you try to evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ by direct substitution:

$$\frac{2^2 - 4}{2 - 2} = \frac{0}{0}$$

This is called an indeterminate form. The expression $\frac{0}{0}$ does not mean the limit is zero, nor does it mean the limit does not exist. It simply means “more work is needed.” The fraction $\frac{0}{0}$ is indeterminate because you cannot determine the limit from that form alone; you need to simplify the expression first.

Think of it this way: $\frac{0}{0}$ is a signal that the numerator and denominator share a common factor that is causing both to hit zero. Your job is to find and cancel that factor.

Factoring to Resolve Indeterminate Forms

The most common technique for dealing with $\frac{0}{0}$ is to factor the numerator and denominator, then cancel the common factor.

Returning to our example:

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

Notice that $x^2 - 4$ is a difference of squares: $x^2 - 4 = (x+2)(x-2)$.

$$\lim_{x \to 2} \frac{(x+2)(x-2)}{x - 2}$$

Now cancel the $(x - 2)$ terms (this is valid because we are taking a limit as $x$ approaches 2, not evaluating at exactly $x = 2$):

$$\lim_{x \to 2} (x + 2) = 2 + 2 = 4$$

The limit is 4. The function $\frac{x^2 - 4}{x - 2}$ has a hole at $x = 2$, but the limit tells us what value the function approaches as $x$ gets close to 2.

Rationalizing Techniques: Using Conjugates

When your expression involves square roots and direct substitution gives $\frac{0}{0}$, factoring might not be obvious. Instead, try rationalizing by multiplying by the conjugate.

The conjugate of an expression like $\sqrt{x} - 3$ is $\sqrt{x} + 3$. The conjugate simply has the opposite sign between the terms. When you multiply an expression by its conjugate, you use the difference of squares pattern to eliminate the square root:

$$(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9$$

This technique is especially useful for limits like:

$$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$$

Direct substitution gives $\frac{0}{0}$. Multiply numerator and denominator by the conjugate of the numerator:

$$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \lim_{x \to 9} \frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$$

Cancel the $(x - 9)$ terms:

$$\lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$$

Limits of Piecewise Functions

A piecewise function is defined by different formulas on different intervals. To evaluate the limit at a boundary point (where the formula changes), you need to check the one-sided limits:

• The left-hand limit $\lim_{x \to a^-} f(x)$ uses the formula that applies when $x < a$.
• The right-hand limit $\lim_{x \to a^+} f(x)$ uses the formula that applies when $x > a$.

The overall limit $\lim_{x \to a} f(x)$ exists only if both one-sided limits exist and are equal.

For example, consider:

$$f(x) = \begin{cases} x^2 & \text{if } x < 2 \ 3x - 2 & \text{if } x \geq 2 \end{cases}$$

To find $\lim_{x \to 2} f(x)$:

• Left-hand limit: $\lim_{x \to 2^-} x^2 = 4$
• Right-hand limit: $\lim_{x \to 2^+} (3x - 2) = 3(2) - 2 = 4$

Since both one-sided limits equal 4, the overall limit is $\lim_{x \to 2} f(x) = 4$.

The Squeeze Theorem: When Direct Methods Fail

Sometimes you encounter functions that oscillate wildly or behave erratically in ways that make factoring or rationalizing impossible. The Squeeze Theorem (also called the Sandwich Theorem) handles these cases by trapping the difficult function between two simpler ones.

Squeeze Theorem: If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (except possibly at $a$ itself), and if:

$$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$$

then:

$$\lim_{x \to a} f(x) = L$$

The intuition is simple: if you squeeze something between two things that both approach the same value, the thing in the middle has no choice but to approach that value too.

A classic example is $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$. The function $\sin\left(\frac{1}{x}\right)$ oscillates wildly between $-1$ and $1$ as $x$ approaches 0. But we know:

$$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$

Multiplying by $x^2$ (which is always non-negative):

$$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$

Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, the Squeeze Theorem tells us:

$$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$$

Notation and Terminology

Term	Meaning	Example
Direct substitution	Evaluate the limit by plugging in the value directly	$\lim_{x \to 2} x^2 = 4$
Indeterminate form	Result like $\frac{0}{0}$ that requires further work	$\lim_{x \to 2} \frac{x^2-4}{x-2}$ gives $\frac{0}{0}$
Limit laws	Rules for breaking limits into simpler pieces	Sum, product, quotient laws
Conjugate	Expression with opposite sign between terms	Conjugate of $\sqrt{x}-3$ is $\sqrt{x}+3$
One-sided limit	Limit approaching from one direction only	$\lim_{x \to 2^-} f(x)$ approaches from the left
Squeeze Theorem	If $g(x) \leq f(x) \leq h(x)$ and limits of $g,h$ are equal, so is limit of $f$	Useful for oscillating functions
Continuous function	Function with no holes, jumps, or asymptotes at a point	Polynomials are continuous everywhere

Examples

Key Properties and Rules

When Direct Substitution Works

Direct substitution works for:

• Polynomial functions at any point
• Rational functions at points where the denominator is not zero
• Root functions at points in their domain
• Trigonometric functions at points where they are defined
• Any function that is continuous at the point in question

Steps for Evaluating Limits Algebraically

1. Try direct substitution first. If you get a real number, you are done.
2. If you get $\frac{0}{0}$, the limit probably exists but requires algebraic simplification:
   • Look for common factors to cancel (try factoring)
   • If square roots are involved, try multiplying by the conjugate
   • Simplify until direct substitution works
3. If you get $\frac{c}{0}$ where $c \neq 0$, the limit is likely infinite or does not exist. Check one-sided limits.
4. For piecewise functions, evaluate left-hand and right-hand limits separately.
5. For functions that oscillate or behave wildly, consider the Squeeze Theorem.

Important Limit Values to Know

These limits appear frequently and are worth memorizing:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

$$\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$

These are all indeterminate forms of type $\frac{0}{0}$, but their values have been established and can be used as building blocks for more complex limits.

Limit Laws Summary

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

Operation	Rule
Sum	$\lim_{x \to a} [f(x) + g(x)] = L + M$
Difference	$\lim_{x \to a} [f(x) - g(x)] = L - M$
Product	$\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
Quotient	$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, if $M \neq 0$
Power	$\lim_{x \to a} [f(x)]^n = L^n$
Constant	$\lim_{x \to a} c = c$

Real-World Applications

Instantaneous Rates Before Derivatives

The entire concept of the derivative is built on limits. When you want to find the instantaneous velocity of a moving object, you compute:

$$\lim_{h \to 0} \frac{f(t + h) - f(t)}{h}$$

This is the slope of the tangent line, found by taking the limit of slopes of secant lines. Evaluating limits algebraically is the technical skill that makes derivatives possible.

Behavior at Critical Points

Engineers and scientists often need to know how systems behave at specific values, including values where the formula might break down. For example, in physics, certain equations for forces or fields become undefined at specific points (like the center of a charged sphere). Limits let you analyze what happens as you approach these critical points, even when you cannot evaluate directly at them.

Error Analysis and Tolerances

In manufacturing and engineering, you often want to know: “If I am slightly off from the target value, how much error will that introduce?” This is fundamentally a question about limits. If a function has a limit at a point, small deviations in input lead to small deviations in output, and you can quantify the tolerances needed.

Compound Interest and Continuous Growth

The formula for continuous compound interest involves the limit:

$$\lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^n = e^r$$

This limit transforms the discrete process of compounding (daily, hourly, every second) into continuous growth. Understanding how to evaluate limits like this reveals why exponential functions are so fundamental to modeling growth.

Signal Processing and Approximations

In signal processing and applied mathematics, functions are often approximated by simpler ones that agree in their limiting behavior. The accuracy of these approximations depends on limits, making algebraic limit evaluation essential for understanding error bounds in approximations.

Self-Test Problems

Summary

• Direct substitution is your first and best tool: if the function is continuous at a point, just plug in the value.
• Limit laws let you break complicated limits into simpler pieces (sums, products, quotients, powers).
• The form $\frac{0}{0}$ is indeterminate, meaning more work is needed, not that the limit fails to exist.
• Factoring is the primary technique for resolving $\frac{0}{0}$ forms: factor, cancel the common factor, then substitute.
• Rationalizing with conjugates works when square roots create indeterminate forms: multiply by the conjugate to eliminate the root.
• For piecewise functions, check both one-sided limits at boundary points; the overall limit exists only if they match.
• The Squeeze Theorem handles wild functions by trapping them between two simpler functions with known limits.
• Algebraic limit evaluation is the technical foundation for derivatives and all of calculus that follows.
• When in doubt, always try direct substitution first. If that fails, look for factors to cancel or conjugates to multiply.