Graph Description: A secant line passes through two points on the curve \(y = f(x)\): point \((a, f(a))\) and point \((b, f(b))\). The slope represents the average rate of change between these points.
Secant Line Slope: The slope of the line through two points \((a, f(a))\) and \((b, f(b))\) on the graph of \(y = f(x)\) is:
\[\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{f(b) - f(a)}{b - a}\]Suppose \(f(x)\) represents distance at time \(t = x\).
Average Velocity: The average velocity on the interval \([a, b]\) is:
\[\text{Average velocity} = \frac{f(b) - f(a)}{b - a}\]If \(s(t) = t^2\), find the average velocity from \(t = 2\) to \(t = 3\).
Solution:
\[\text{Average velocity} = \frac{s(3) - s(2)}{3 - 2} = \frac{3^2 - 2^2}{1} = \frac{9 - 4}{1} = 5 \text{ mi/hr}\]What is the velocity at an instant?
Key Idea: To find instantaneous velocity at \(t = 2\), make intervals smaller and smaller around \(t = 2\) and find the average velocity.
For \(s(t) = t^2\), let's calculate average velocities on shrinking intervals:
| Interval | Average Velocity (mi/hr) |
|---|---|
| \([2, 3]\) | \(5\) |
| \([2, 2.5]\) | \(4.5\) |
| \([2, 2.1]\) | \(4.1\) |
| \([2, 2.01]\) | \(4.01\) |
| \([2, 2.00001]\) | \(4.00001\) |
From the left:
| Interval | Average Velocity (mi/hr) |
|---|---|
| \([1.9, 2]\) | \(3.9\) |
| \([1.99, 2]\) | \(3.99\) |
| \([1.999, 2]\) | \(3.999\) |
Conclusion: The instantaneous velocity at \(t = 2\) is \(4\) mi/hr.
Limit Definition: The idea of finding the value of something when you get closer to another quantity.
Notation: \(\lim_{x \to a} f(x)\)
Reads as: "The limit of \(f(x)\) as \(x\) approaches \(a\)"
Meaning: The value of \(f(x)\) when \(x\) gets closer to \(a\)
From our example: Instantaneous velocity at \(t = 2\) is \(\lim_{t \to 2}\) [average velocity]
Note: When \(x = 3\), both numerator and denominator equal 0, so \(f(3)\) is not defined.
However, \(\lim_{x \to 3} f(x)\) may still exist for \(x \neq 3\).
| \(x\) (approaching from right) | \(f(x)\) |
|---|---|
| \(3.01\) | \(6.01\) |
| \(3.001\) | \(6.001\) |
| \(3.00001\) | \(6.00001\) |
| \(x\) (approaching from left) | \(f(x)\) |
|---|---|
| \(2.99\) | \(5.99\) |
| \(2.999\) | \(5.999\) |
| \(2.9999\) | \(5.9999\) |
Conclusion: \(\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6\)
Graph Description: The graph shows \(f(x) = x + 3\) with a hole at \(x = 3\). As \(x\) approaches 3 from both the right and left, the function values approach 6, even though \(f(3)\) is not defined.
From the graph:
Conclusion: \(\lim_{x \to 3} f(x) = 6\)
Graph Description: The graph shows a piecewise function with a jump discontinuity at \(x = -2\) and different behavior at \(x = 3\).
At \(x = -2\):
At \(x = 3\):
Graph Description: The graph shows different behaviors at \(x = 1\) and \(x = 5\), with the function approaching different values from the left and right at \(x = 1\).
At \(x = 1\):
At \(x = 5\):
Right-Hand Limit: \(\lim_{x \to a^+} f(x)\)
The function value as \(x\) approaches \(a\) from the right
Left-Hand Limit: \(\lim_{x \to a^-} f(x)\)
The function value as \(x\) approaches \(a\) from the left
Theorem: \(\lim_{x \to a} f(x) = L\) if and only if:
\[\lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L\]If the left and right limits are different, then \(\lim_{x \to a} f(x) = \text{DNE}\)
From our previous examples:
Key Point: A limit exists at a point if and only if the left-hand and right-hand limits exist and are equal. The actual function value at that point is irrelevant for the existence of the limit.