Lesson 6 (09/04/20)

Today:

Infinite limits
Vertical asymptotes

Office Hours: Monday, Wednesday, Friday: 130pm-230pm

Announcements:

Exam 1 on Wednesday 09/23, 8pm-9pm
HW 6 Due: Tuesday 09/08
Quiz 3 (Lesson 5): Tuesday 09/08

Warmup: evaluate the following:

\[ \frac{5}{0.1} = 50 \]

\[ \frac{5}{0.01} = 500 \]

\[ \frac{5}{0.001} = \]

\[ \frac{5}{0.0001} = \]

\[ \frac{5}{0.000001} = 5000000 \]

\[ \frac{\text{Number}}{\text{small +ve}} = \text{LAARGE } +\text{VE} \]

\[ +\infty \]

\[ \frac{93}{-0.01} = -93 \]

\[ \frac{93}{-0.001} = -930 \]

\[ \frac{93}{-0.0001} = \]

\[ \frac{93}{-0.000001} = \]

\[ \frac{93}{-0.00000001} = -9300000000 \]

\[ \frac{\text{Number}}{\text{small } -\text{ve Number}} = \text{LAARGE } -\text{ve Number} \]

\[ -\infty \]

Infinite Limits

eg: \[ \lim_{x \to 7} \frac{1}{x-7} \]

\( x \) close to 7, but bigger

\[ x = 7.1 \implies \frac{1}{x-7} = \frac{1}{0.1} = 10 \] \[ x = 7.01 \implies \frac{1}{7.01-7} = \frac{1}{0.01} = 100 \] \[ x = 7.0001 \implies \frac{1}{7.0001-7} = \frac{1}{0.0001} = 10000 \] \[ x = 7.0000001 \implies \frac{1}{7.0000001-7} = 10000000. \]

\( \frac{1}{x-7} \) is Getting LAARGE +ve

\( x \) close to 7, but bigger

\[ x = 6.9 \implies \frac{1}{x-7} = -10 \] \[ x = 6.999 \implies \frac{1}{x-7} = -1000 \] \[ x = 6.99999 \implies \frac{1}{x-7} = -100000 \] \[ x = 6.999999 \implies \frac{1}{x-7} = -10000000 \]

\( \frac{1}{x-7} \) is getting LAARGE -ve

Graph of \( f(x) = \frac{1}{x-7} \), near \( x = 7 \)

A Cartesian coordinate system illustrating the behavior of the function f(x) = 1/(x - 7) near x = 7. A vertical dashed line marks the line x = 7. To the right, as x approaches 7 from the right (indicated by an arrow along the x-axis pointing left toward 7), the curve ascends upward towards positive infinity (+∞). To the left, as x approaches 7 from the left (indicated by an arrow along the x-axis pointing right toward 7), the curve descends downward towards negative infinity (-∞).

Visual Description: A Cartesian coordinate system illustrating the behavior of the function f(x) = 1/(x - 7) near x = 7. A vertical dashed line marks the line x = 7. To the right, as x approaches 7 from the right (indicated by an arrow along the x-axis pointing left toward 7), the curve ascends upward towards positive infinity (+∞). To the left, as x approaches 7 from the left (indicated by an arrow along the x-axis pointing right toward 7), the curve descends downward towards negative infinity (-∞).

\[ \lim_{x \to 7^+} \frac{1}{x-7} = +\infty \]

\[ \lim_{x \to 7^-} \frac{1}{x-7} = -\infty \]

\[ \lim_{x \to 7} \frac{1}{x-7} \text{ DNE} \]

\( x = 7 \) is called vertical Asymptote

Given \(f(x)\)

\(x = a\) is called a vertical asymptote if

\[ \lim_{x \to a^+} f(x) = \pm \infty \quad \text{OR} \quad \lim_{x \to a^-} f(x) = \pm \infty \]

Example 1

\[ \lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 7} \]

If you try to plug in number close to \(x = 7\) in the denominator you get a small number in the denominator.

Do NOT mean you have an Asymptote

\[ \lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 7} = \lim_{x \to 7} \frac{(x - 7)(x + 3)}{(x - 7)} = \lim_{x \to 7} (x + 3) = 10 \]

You have a hole in the Graph.

\[ \frac{1}{x-7} \]

\(x = 7\) is V.A

Cartesian coordinate plot of the rational function f(x) = 1/(x-7). A vertical dashed line indicates the vertical asymptote at x = 7. The curve approaches negative infinity as x approaches 7 from the left, passing through the fourth quadrant with a downward arrow, and approaches positive infinity as x approaches 7 from the right with an upward arrow in the first quadrant.

Visual Description: Cartesian coordinate plot of the rational function f(x) = 1/(x-7). A vertical dashed line indicates the vertical asymptote at x = 7. The curve approaches negative infinity as x approaches 7 from the left, passing through the fourth quadrant with a downward arrow, and approaches positive infinity as x approaches 7 from the right with an upward arrow in the first quadrant.

\[ \frac{x^2-4x-21}{x-7} = \frac{(x+3)(x-7)}{(x-7)} \]

Cartesian coordinate plot illustrating the function f(x) = (x^2 - 4x - 21) / (x - 7). The graph is a line with a positive slope representing y = x + 3, with a removable discontinuity (a hole represented by an open circle) at the point (7, 10). Dashed guide lines connect the hole to 7 on the x-axis and 10 on the y-axis.

Visual Description: Cartesian coordinate plot illustrating the function f(x) = (x^2 - 4x - 21) / (x - 7). The graph is a line with a positive slope representing y = x + 3, with a removable discontinuity (a hole represented by an open circle) at the point (7, 10). Dashed guide lines connect the hole to 7 on the x-axis and 10 on the y-axis.

Example 1: \( \lim_{x \to 7} \frac{x^2 - 4x - 20}{x - 7} \)

if we plug in \( 7 \to \frac{7^2 - 4(7) - 20}{7 - 7} = \frac{1}{0} \quad \text{Actually } \frac{1}{\text{Small}} \)

\[ \lim_{x \to 7^+} \frac{x^2 - 4x - 20}{x - 7} = \frac{\approx 1}{\text{Small +ve}} = \text{LARGE +VE} = +\infty \]

\[ \lim_{x \to 7^-} \frac{x^2 - 4x - 20}{x - 7} = \frac{\approx 1}{\text{Small -ve}} = \text{LARGE -VE} = -\infty \]

\[ \lim_{x \to 7} \frac{x^2 - 4x - 20}{x - 7} \quad \text{DNE} \]

\( x = 7 \) is a Vertical Asymptote.

Example 2

\[ \lim_{x \to 7} \frac{1}{(x-7)^2} \approx \frac{1}{\text{small } +\text{ve}} = \frac{\text{LARGE}}{+\text{ve}} = +\infty \]

\[ \lim_{x \to 7^-} \frac{1}{(x-7)^2} \quad \text{is} \quad \frac{1}{(-0.1)^2}, \frac{1}{(-0.001)^2} \dots \quad \frac{\text{LARGE}}{+\text{ve}} = +\infty \]

\[ \lim_{x \to 7^+} \frac{1}{(x-7)^2} = \frac{\text{LARGE}}{+\text{ve}} = +\infty \]

Hand-drawn graph of y = 1 / (x - 7)^2 on Cartesian coordinates. A dashed vertical line is drawn at x = 7 representing a vertical asymptote. As x approaches 7 from the left, the curve rises steeply upward toward positive infinity with an upward arrow. As x approaches 7 from the right, the curve similarly ascends toward positive infinity with an upward arrow. The curve crosses the y-axis above the x-axis.

Visual Description: Hand-drawn graph of y = 1 / (x - 7)^2 on Cartesian coordinates. A dashed vertical line is drawn at x = 7 representing a vertical asymptote. As x approaches 7 from the left, the curve rises steeply upward toward positive infinity with an upward arrow. As x approaches 7 from the right, the curve similarly ascends toward positive infinity with an upward arrow. The curve crosses the y-axis above the x-axis.

\(x = 7\) is a vertical asymptote

\[ \lim_{x \to 7} \frac{1}{(x-7)^2} = +\infty \]

eg ③

\[ \lim_{x \to 7^+} \frac{x+11}{\sqrt{x-7}} \]

\[ \lim_{x \to 7^-} \frac{x+11}{\sqrt{x-7}} \quad \text{DNE} \quad \text{bcz } x < 7 \text{ is not in domain.} \]

but when \( x > 7 \)

\[ \lim_{x \to 7^+} \frac{x+11}{\sqrt{x-7}} \approx \frac{18}{\sqrt{\text{small Number}}} = \frac{18}{\begin{matrix} \text{small} \\ \text{+ve} \\ \text{Number} \end{matrix}} = +\infty, \]

\[ \sqrt{0.00000001} = \underset{\text{small}}{0.0001} \]

\( x = 7 \) is a V.A.

Example 4: Evaluate \(\lim_{x \to -1} \frac{x^2 - 4x + 3}{x^2 - 1}\)

\(x = -1 \implies N: (-1)^2 - 4(-1) + 3 = 8\)
\(D: (-1)^2 - 1 = 0\)

\[ \frac{x^2 - 4x + 3}{x^2 - 1} \approx \frac{8}{\text{small}} = \text{LARGE} \] \[ \lim_{x \to -1^+} \frac{x^2 - 4x + 3}{x^2 - 1} \approx \frac{8}{\begin{matrix} \text{small} \\ -\text{ve} \end{matrix}} = -\infty \] \[ \lim_{x \to -1^-} \frac{x^2 - 4x + 3}{x^2 - 1} \approx \frac{8}{\begin{matrix} \text{small} \\ +\text{ve} \end{matrix}} = +\infty \] \[ \lim_{x \to -1} \frac{x^2 - 4x + 3}{x^2 - 1} \text{ DNE} \]

\(x = -1\) is a V.A

Example 5: Evaluate \( \lim_{x \to 1} \frac{x^2 - 4x + 3}{x^2 - 1} \)

\( x = 1 \) in

\( N: 1^2 - 4(1) + 3 = 0 \)
\( D: 1^2 - 1 = 0 \)

\[ \frac{0}{0} \]

\( \underline{N}: x^2 - 4x + 3 = (x - 1)(x - 3) \)
\( D: x^2 - 1 = (x - 1)(x + 1) \)

\[ \lim_{x \to 1} \frac{x^2 - 4x + 3}{x^2 - 1} = \lim_{x \to 1} \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{(x - 3)}{x + 1} = \frac{-2}{2} = -1 \]

at \( x = 1 \) you have a hole.

Example 6

\[ f(x) = \frac{x^2 - 3x - 28}{(x - a)(x - 1)} \]

For what values of \( a \), \( f(x) \) has a vertical asymptote at \( x = a \)

Want:
Numerator \( \neq 0 \)
Denominator \( = 0 \)
when you plug in \( x = a \)

Want Numerator \( \neq 0 \) when you plug in \( x = a \)
that means \( x - a \) is not a factor of Numerator

\( N: x^2 - 3x - 28 = (x + 4)(x - 7) \)

as long as \( a \neq -4 \), \( a \neq 7 \), Numerator \( \neq 0 \).

\( \Rightarrow \) Have a V.A at \( x = a \), for all values \( a \), except \( -4 \) & \( 7 \).