\( y = f(x) \)
The line that goes through the points \( (a, f(a)) \) and \( (b, f(b)) \) is called the secant line whose slope is:
\[ \frac{f(b) - f(a)}{b - a} \]
If \( f(x) \) measures distance and \( x \) is time, then the secant line slope is the average velocity over the interval \( a \) to \( b \).
If \( s(t) = t^2 \) gives us the distance of an object from the origin on a straight line:
distance: \( s(t) = t^2 \)
On average, how fast did the object move between \( t = 2 \) and \( t = 3 \)?
\[ \text{Average velocity: } \frac{s(3) - s(2)}{3 - 2} = \frac{9 - 4}{1} = 5 \]
On average, the object moved at a velocity of 5 during \( 2 \le t \le 3 \).
But that does NOT mean it traveled at a velocity of 5 at all times. The instantaneous velocity is always changing.
What if we want the instantaneous velocity at a specific time? For example, at \( t = 2 \)?
To find the instantaneous velocity at \( t = 2 \), we calculate the average velocity using smaller and smaller time intervals.
For example, \( 2 \le t \le 2.5 \)
Interval is so short, average velocity is practically the instantaneous velocity.
When \( b \) is practically on top of \( a \), secant line \( \to \) tangent line.
Slope = average velocity on \( [a, b] \)
By moving \( b \) closer and closer to \( a \), the secant line slope approach the instantaneous velocity at \( t = a \).
The idea of bringing something closer to another is basically how limit in calculus works.
We write: \( \lim_{x \to a} f(x) \)
Means we want to know what \( f(x) \) gets close to as \( x \) gets close to \( a \).
Make \( x \) close to \( a \) without equalling \( a \).
\( \lim_{x \to 10} x^2 \) means find what \( x^2 \) gets close to as \( x \) gets close to 10.
| \( x \) | 9.9 | 9.999 | 10 | 10.001 | 10.1 |
|---|---|---|---|---|---|
| \( f(x) = x^2 \) | 98.01 | 99.98001 | X | 100.02 | 102.01 |
The bottom row suggests \( \lim_{x \to 10} x^2 = 100 \).
The fact that \( x \) can be 10 is irrelevant for limit.
another example :
Clearly, we cannot go to exactly \( x = -4 \) but we can get close to \( x = -4 \). And that is what the limit is about.
| \( x \) | -4.01 | -4.001 | -4.0001 | -4 | -3.9999 | -3.999 | -3.99 |
|---|---|---|---|---|---|---|---|
| \( \frac{x^2-16}{x+4} \) | -8.01 | -8.001 | -8.0001 | X | -7.9999 | -7.999 | -7.99 |
So the bottom row suggests that \( \frac{x^2-16}{x+4} \) "wants" to be -8 as \( x \) gets close to -4.
looks like a line with a point removed at \( x = -4 \)
we can make \( \frac{x^2-16}{x+4} \) as close to -8 as we want by making \( x \) close to -4
The following graph illustrates the behavior of a function at specific points to demonstrate the difference between function values and limits.
Note on the graph: at \( x = 3 \), \( y \to 6 \) no matter how we approach \( x = 3 \). At \( x = 5 \), coming from the left, \( y \to -4 \), and coming from the right, \( y \to -7 \).
\( f(3) \) does not exist (DNE) because of the open circle
but \[ \lim_{x \to 3} f(x) = 6 \]
\( f(5) = -4 \to \) we can get to \( x = 5 \)
but \[ \lim_{x \to 5} f(x) \text{ does not exist because the } y \text{ values approach different numbers depending on which side we come from} \]
Even though \( \lim_{x \to 5} f(x) \) does not exist, because the value depends on how we approach \( x = 5 \), if we only look at one side at a time, we can still have one-sided limits.
\[ \lim_{x \to 5^-} f(x) = -4 \]
Approach \( x = 5 \) from the LEFT
\[ \lim_{x \to 5^+} f(x) = -7 \]
Approach \( x = 5 \) from the RIGHT
If these are different, then the limit at \( x = 5 \) does not exist.