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Secant line = line through two points on graph
\[ \text{Slope} = \frac{\text{change in } y\text{-coordinate}}{\text{change in } x\text{-coordinate}} \]
\[ = \frac{f(b) - f(a)}{b - a} \]
if \(f(x)\) represents distance then \(\frac{f(b)-f(a)}{b-a}\) is Average velocity on \([a, b]\)
Eg: \( S(t) = t^2\text{ miles} \), find average velocity from \( t = 2\text{ hr} \) to \( t = 3\text{ hr} \)
Average velocity
\[ = \frac{S(3) - S(2)}{3 - 2} \] \[ = \frac{9 - 4}{1} = 5\text{ mi/hr.} \]
What is velocity at \( \underline{\underline{t = 2}} \) \( \Big\} \) instantaneous velocity at \( t = 2 \)
idea: find Average velocity in smaller intervals
\( S(t) = t^2 \), Average velocity on \( [a, b] \) is
\[ \frac{S(b) - S(a)}{b - a} = \frac{b^2 - a^2}{b - a} = \frac{\cancel{(b - a)}(b + a)}{\cancel{(b - a)}} = b + a \]
on \( [2, 2.1] \) Average Velocity is \( 2 + 2.1 = 4.1 \)
\( [2, 2.01] \) Average Velocity is \( 4.01 \)
\( [2, 2.001] \) Average Velocity is \( 4.001 \)
\( [1.9, 2] \) Average Velocity is \( 3.9 \)
\( [1.99, 2] \) Average Velocity is \( 3.99 \)
\( [1.999, 2] \) Average Velocity is \( 3.999 \)
As interval gets smaller, Av. velocity gets closer to \( 4 \)
\( \rightsquigarrow \text{instantaneous velocity} = 4\text{ mi/hr} \)
Value of something [function value / output] when something else [input] gets closer to a given number.
Notation: \(\lim_{x \to a} f(x)\)
Previous example:
Av. velocity on \([2, x] = \frac{S(x) - S(2)}{x - 2}\)
instantaneous velocity at \(x = 2 = \lim_{x \to 2} \frac{S(x) - S(2)}{x - 2} = 4\)
Numerically find value of \( f(x) = \frac{x^2 - 9}{x - 3} \) when \( x \) is close to \( 3 \).
when \( x \neq 3 \leadsto f(x) = \frac{(x+3)\cancel{(x-3)}}{\cancel{(x-3)}} = x + 3 \).
\( f(3.1) = 3.1 + 3 = 6.1 \)
\( f(3.01) = 3.01 + 3 = 6.01 \)
\( f(3.0000001) = 3.0000001 + 3 = 6.0000001 \)
\( f(2.9) = 5.9 \)
\( f(2.999) = 5.999 \)
\( f(2.9999999) = 5.9999999 \)
as \( x \) gets closer to \( 3 \), \( f(x) \) gets closer to \( 6 \)
\[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} f(x) = 6. \]
\[ f(x) = \frac{x^2-9}{x-3} \leadsto \text{Domain} = (-\infty, 3) \cup (3, \infty) \] \[ = \frac{(x-3)(x+3)}{x-3} \] \[ f(x) = \begin{cases} x+3 & x \neq 3 \\ \text{Not defined} & x = 3 \end{cases} \]
\( f(x) \) approach \( 6 \) as \( x \) approach \( 3 \) from left
\( f(x) \) approach \( 6 \) as \( x \) approach \( 3 \) from Right
\[ \left. \begin{aligned} & f(x) \text{ approach } 6 \text{ as } x \text{ approach } 3 \text{ from left} \\ & f(x) \text{ approach } 6 \text{ as } x \text{ approach } 3 \text{ from Right} \end{aligned} \right\} \implies \lim_{x \to 3} f(x) = 6 \]
eg: evaluate \( \lim_{x \to -2} f(x) \) and \( \lim_{x \to 3} f(x) \)
\( f(x) \) Approach 4 from both right & left as \( x \) approach 3
\[ \lim_{x \to 3} f(x) = 4 \]
Despite \( f(3) = 5 \)
\( f(x) \) approach 2 as from both sides as \( x \) approach -2
\[ \lim_{x \to -2} f(x) = 2 \]
\( f(x) \) approach \(-4\) as \( x \) approach \( 5 \) from Right
Also approach \(-4\) as \( x \) approach \( 5 \) from left
\[ \lim_{x \to 5} f(x) = -4 \]
\( f(x) \) approach \( 2 \) as \( x \) approach \( 1 \) from Right
\( f(x) \) approach \( 4 \) as \( x \) approach \( 1 \) from left
\[ 2 \neq 4 \]
\( \rightsquigarrow \lim_{x \to 1} f(x) \) Does not exist
\( \lim_{x \to a^-} f(x) \) is called left limit
= value of \( f(x) \) as \( x \) approach \( a \) from left
\[ \lim_{x \to 1^-} f(x) = 4 \]
\[ \lim_{x \to 5^-} f(x) = -4 \]
\( \lim_{x \to a^+} f(x) \) is called Right limit
= value of \( f(x) \) as \( x \) approach \( a \) from Right
\[ \lim_{x \to 1^+} f(x) = 2 \]
\[ \lim_{x \to 5^+} f(x) = -4 \]
Suppose
\[ \lim_{x \to a^+} f(x) = L \]
and
\[ \lim_{x \to a^-} f(x) = L \]
then
\[ \lim_{x \to a} f(x) = L \]
Suppose
\[ \lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x) \]
then
\[ \lim_{x \to a} f(x) \]
Does not exist