Review: One-Sided and Two-Sided Limits
The concept of a limit is fundamental in calculus, describing a function's behavior as its input approaches a specific point. The distinction between one-sided and two-sided limits is crucial for this analysis, especially for functions with breaks or "jumps."
One-Sided Limits
A one-sided limit examines the function's value as it approaches a point from a single direction. This is particularly useful for analyzing functions with "jumps" or breaks.
- The left-hand limit (\( \lim_{x \to c^-} f(x) \)) examines the function's behavior as \(x\) approaches \(c\) from values less than \(c\).
- The right-hand limit (\( \lim_{x \to c^+} f(x) \)) looks at the function's behavior as \(x\) approaches \(c\) from values greater than \(c\).
One-sided limits are essential for analyzing piecewise functions, where the function's definition changes at a specific point. For example, the function \(f(x) = |x|/x\) has a left-hand limit of \(-1\) at \(x=0\) and a right-hand limit of \(1\).
---Two-Sided Limits and the Condition for Existence
A two-sided limit, or simply "the limit," exists at a point \(c\) if and only if both the left-hand and right-hand limits at that point exist and are equal to the same finite value.
\[ \lim_{x \to c} f(x) = L \quad \text{if and only if} \quad \lim_{x \to c^-} f(x) = L \quad \text{and} \quad \lim_{x \to c^+} f(x) = L \]If the two one-sided limits are different, the two-sided limit does not exist (DNE). This is because for a limit to exist, the function must be consistently approaching a single value from both sides, like a road converging to a single point. If the roads from the left and right lead to different places, there's no single limit point. This concept is at the heart of understanding discontinuities in a function.
---Check Yourself Problems
Test your understanding of one-sided and two-sided limits. For each problem, determine if the limit exists and, if so, what its value is.
Problem 1: Given the function \(f(x) = \begin{cases} x+2 & \text{if } x < 1 \\ 3 & \text{if } x \ge 1 \end{cases}\), find \(\lim_{x \to 1} f(x)\).
First, find the one-sided limits:
- Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 1+2 = 3 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 3 = 3 \)
Since the left-hand limit equals the right-hand limit (both are 3), the two-sided limit exists and is 3. \( \lim_{x \to 1} f(x) = 3 \).
Problem 2: Find \( \lim_{x \to 5} \frac{x^2 - 3x - 10}{x - 5} \).
This is an indeterminate form \( \frac{0}{0} \), so we factor the numerator to simplify:
\[ \lim_{x \to 5} \frac{(x-5)(x+2)}{x-5} = \lim_{x \to 5} (x+2) = 5+2 = 7 \]The limit is 7.
Problem 3: Find \( \lim_{x \to 0} \frac{\sin(4x)}{x} \).
This is a fundamental trigonometric limit. To use the identity \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \), we need to manipulate the expression so the argument of the sine function matches the denominator. We can multiply the numerator and denominator by 4:
\[ \lim_{x \to 0} \frac{4 \sin(4x)}{4x} = 4 \left( \lim_{x \to 0} \frac{\sin(4x)}{4x} \right) = 4(1) = 4 \]The limit is 4.
Problem 4: Find \( \lim_{x \to 4^-} \frac{x+5}{x-4} \).
As \(x\) approaches 4 from the left, the numerator approaches \(4+5=9\), which is a positive number. The denominator, \(x-4\), approaches 0 from the negative side (e.g., \(3.9-4 = -0.1\)).
A positive number divided by a very small negative number results in a very large negative number. Therefore, the limit is \(-\infty\).
\[ \lim_{x \to 4^-} \frac{x+5}{x-4} = -\infty \]Problem 5: Find \( \lim_{x \to 3^+} \frac{1}{x-3} \).
As \(x\) approaches 3 from the right (e.g., 3.1), the numerator is a constant 1. The denominator, \(x-3\), approaches 0 from the positive side (e.g., \(3.1-3=0.1\)).
A positive number divided by a very small positive number results in a very large positive number. Therefore, the limit is \(+\infty\).
\[ \lim_{x \to 3^+} \frac{1}{x-3} = +\infty \]