1.2: The Limit Laws — Read — Invisible Network

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Section 1.2 The Limit Laws

Learning Objectives

Evaluate the limit of a function using the limit laws.
Evaluate the limit of a function at a specific point by factoring the expression or using conjugates.
Apply the squeeze theorem to evaluate the limit of a function at a point where direct substitution leads to an indeterminate form.

Suppose that \( \displaystyle{\lim_{x\to a } f(x) =L} \) and \( \displaystyle{\lim_{x\to a } g(x)=M} \) exist and that \( c \) is a constant.

(a) \( \displaystyle{\lim_{x\to a } [f(x)+g(x)] = \lim_{x\to a } f(x)+\lim_{x\to a } g(x) = L+M} \)

(b) \( \displaystyle{\lim_{x\to a } [f(x)-g(x)] = \lim_{x\to a } f(x) - \lim_{x\to a } g(x) = L-M} \)

(d) \( \displaystyle{\lim_{x\to a } [f(x)g(x)] = \lim_{x\to a } f(x) \cdot \lim_{x\to a } g(x) = L \cdot M} \)

(e) \( \displaystyle{\lim_{x\to a } \frac{f(x)}{g(x)} = \frac{\lim_{x\to a } f(x)}{\lim_{x\to a } g(x)} = \frac{L}{M}} \), provided \( \displaystyle{\lim_{x\to a } g(x) \neq 0} \)

(f) \( \displaystyle{\lim_{x\to a } [f(x)]^n = \left( \lim_{x\to a } f(x) \right)^n = L^n} \), for any positive integer \( n \).

(g) \( \displaystyle{\lim_{x\to a } \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a } f(x)} = \sqrt[n]{L}} \), for all \( L \) if \( n \) is odd, and for \( L \geq 0 \) if \( n \) is even and \( f(x) \geq 0 \).

(h) \( \displaystyle{\lim_{x\to a } c = c} \), \( \displaystyle{\lim_{x\to a } x = a} \), \( \displaystyle{\lim_{x\to a } x^n = a^n} \), \( \displaystyle{\lim_{x\to a } \sqrt[n]{x} = \sqrt[n]{a}} \) (when defined).

Example \(\PageIndex{11}\)

Use the limit laws to evaluate the following limits: \[ \lim_{x\to -1} (2x+1), \ \lim_{x\to 2} \frac{x^2-3x+4}{x+2} \]

Theorem \(\PageIndex{3}\)

Suppose that \( f \) is a polynomial or rational function. If \( a \) is in the domain of \( f \), then

\[
\lim_{x \to a} f(x) = f(a).
\]

Example \(\PageIndex{12}\)

Use Theorem 1.2.3 to evaluate the following limits: \[ \lim_{x\to -1} (2x+1), \ \lim_{x\to 2} \frac{x^2-3x+4}{x+2} \]

Evaluate the following limit: \(\displaystyle{\lim_{x\to 1} \frac{x-1}{x^2+3x-4}}\).

What happens if you try to factor the numerator? How does this help simplify the expression?
After factoring, what does the new simplified expression suggest about the limit as \(x\) approaches 1?

Techniques to Evaluate Limits

Calculating a limit when \( f(x)/g(x) \) has the indeterminate form \( 0/0 \)

1. Check if the limit can be calculated using the limit laws.

2. Task: Find a function equal to \( h(x) = \dfrac{f(x)}{g(x)} \) for all \( x \neq a \) over some interval containing \( a \).

(a) If \( f(x) \) and \( g(x) \) are polynomials, factor each function and cancel out any common factors.

(b) If the numerator or denominator contains a difference involving a square root, try multiplying the numerator and denominator by the conjugate of the expression involving the square root.

Example \(\PageIndex{13}\)

Evaluate the following limits:

(a) \( \displaystyle{\lim_{x \to 2} \frac{x^2 - 4}{x - 2}} \)

(b) \( \displaystyle{\lim_{x \to 4} \frac{x^2 - 16}{x - 4}} \)

(d) \( \displaystyle{\lim_{x \to 0} \frac{\sqrt{(x^2 + 1)} - 1}{x^2}} \)

(e) \( \displaystyle{\lim_{x \to 2} \frac{\frac{1}{x - 1} - 1}{x - 2}} \)

(f) \( \displaystyle{\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)} \)

(g) \( \displaystyle{\lim_{x \to 4} f(x)} \), where \(
f(x) =
\begin{cases}
\frac{x^2 - 9}{x - 3} & \text{if } x < 4 \\
5 & \text{if } x = 4 \\
x + 2 & \text{if } x > 4
\end{cases}
\)

The Squeeze Theorem

Theorem \(\PageIndex{4}\)

Let \( f(x) \), \( g(x) \), and \( h(x) \) be defined for all \( x \neq a \) over an open interval containing \( a \), and let \( L \) be a real number. If \(
f(x) \leq g(x) \leq h(x) \) for all \( x \neq a \) in an open interval containing \( a \), and

\[
\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x),
\]

then

\[
\lim_{x \to a} g(x) = L.
\]

Example \(\PageIndex{14}\)

Use the limit \( \displaystyle{\lim_{x\to 0} \frac{\sin x}{x} =1} \) to evaluate the limit \[ \displaystyle{\lim_{x\to 0} \frac{1-\cos x}{x} =0} \]

Example \(\PageIndex{15}\)

(a) Let \( f(x) = \begin{cases}
x^2 \sin\left( \dfrac{1}{x} \right), & \text{if } x \neq 0 \\
0, & \text{if } x = 0
\end{cases}
\). Use the Squeeze Theorem to evaluate: \( \displaystyle{\lim_{x \to 0} f(x)} \)

(b) Let \( g(x) =
\begin{cases}
x \cos\left( \frac{1}{x} \right), & \text{if } x \neq 0 \\
0, & \text{if } x = 0
\end{cases}
\). Use the Squeeze Theorem to evaluate: \( \displaystyle{\lim_{x \to 0} g(x)} \)

This page titled 1.2: The Limit Laws was last modified on Sun, 07 Dec 2025 01:10:57 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Kevin Palencia.

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