Lesson 5 (2.3)

Today:

Computing limits
Limit Laws
Squeeze/Sandwich Theorem

Office Hours: Monday, Wednesday, Friday 1:30 pm – 2:30 pm

Announcements:

Quiz 2 (Lessons 3, 4) – Tomorrow (09/03)
Feasting with Faculty – Wednesday's (TODAY)
6pm – 7pm
Wiley Dining Hall

Review example.

A Cartesian coordinate graph illustrating piecewise function limits and continuity at various points. For x < -2, a curve increases to a closed filled circle at (-2, 2). At x = -2, there is an open circle at (-2, 3), from which a straight line segment extends down through the y-intercept at (0, 0.5), marked with a solid dot, ending at an open circle at (1, -1). Below x = 1, there is an isolated filled dot at (1, -2). From the open circle at (1, -1), a curve arches up to an open circle at (3, 1), and then a line slopes downward to the right for x > 3. Dashed lines project key points to their respective x-values (-2, 1, 3) and y-values (-1, 0.5, 1, 2, 3).

Visual Description: A Cartesian coordinate graph illustrating piecewise function limits and continuity at various points. For x < -2, a curve increases to a closed filled circle at (-2, 2). At x = -2, there is an open circle at (-2, 3), from which a straight line segment extends down through the y-intercept at (0, 0.5), marked with a solid dot, ending at an open circle at (1, -1). Below x = 1, there is an isolated filled dot at (1, -2). From the open circle at (1, -1), a curve arches up to an open circle at (3, 1), and then a line slopes downward to the right for x > 3. Dashed lines project key points to their respective x-values (-2, 1, 3) and y-values (-1, 0.5, 1, 2, 3).

At \( x = -2 \):

\[ \lim_{x \to -2^-} f(x) = 2 \]

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

\[ \lim_{x \to -2} f(x) = \text{DNE} \]

\[ f(-2) = 2 \]

At \( x = 0 \):

\[ \lim_{x \to 0^-} f(x) = 0.5 \]

\[ \lim_{x \to 0^+} f(x) = 0.5 \]

\[ \lim_{x \to 0} f(x) = 0.5 \]

\[ f(0) = 0.5 \]

At \( x = 1 \):

\[ \lim_{x \to 1^-} f(x) = -1 \]

\[ \lim_{x \to 1^+} f(x) = -1 \]

\[ \lim_{x \to 1} f(x) = -1 \]

\[ f(1) = -2 \]

At \( x = 3 \):

\[ \lim_{x \to 3^-} f(x) = 1 \]

\[ \lim_{x \to 3^+} f(x) = 1 \]

\[ \lim_{x \to 3} f(x) = 1 \]

\[ f(3) = \text{Not Defined} \]

Finding limits of Simple functions

1. Constant function

eg: \( f(x) = 5 \)

\[ \lim_{x \to a} f(x) = 5 \]

\[ f(x) = c \]

\[ \lim_{x \to a} f(x) = \lim_{x \to a} c = c \]

Cartesian coordinate plane showing a horizontal line representing the constant function f(x) = 5 intersecting the y-axis at 5. A dashed vertical line extends upwards from point a on the x-axis to the horizontal line at (a, 5).

Visual Description: Cartesian coordinate plane showing a horizontal line representing the constant function f(x) = 5 intersecting the y-axis at 5. A dashed vertical line extends upwards from point a on the x-axis to the horizontal line at (a, 5).

2. Straight lines:

\[ f(x) = mx + b \]

\[ \lim_{x \to a} f(x) = \lim_{x \to a} mx + b \]

\[ = ma + b \]

eg:

\[ \lim_{x \to 3} (7x - 5) = 7(3) - 5 \]

\[ = 16 \]

Cartesian coordinate plane displaying the linear function f(x) = mx + b as a straight line with positive slope and positive y-intercept. On the x-axis, horizontal arrows point inward toward x = a from both the left and right. A dashed vertical line rises from x = a to the line f(x) at point (a, ma + b), where arrows along the line also point inward toward the point from both directions, demonstrating the two-sided limit.

Visual Description: Cartesian coordinate plane displaying the linear function f(x) = mx + b as a straight line with positive slope and positive y-intercept. On the x-axis, horizontal arrows point inward toward x = a from both the left and right. A dashed vertical line rises from x = a to the line f(x) at point (a, ma + b), where arrows along the line also point inward toward the point from both directions, demonstrating the two-sided limit.

3. \(f(x) = x^2\)

\[ \lim_{x \to 6} x^2 = 6^2 = 36 \]

Graph of the quadratic function f(x) = x^2 on a set of Cartesian coordinate axes with vertex at the origin (0,0). A point is marked on the right branch of the parabola, with dashed lines extending down to x = 6 on the horizontal axis and left to y = 36 on the vertical axis, illustrating that as x approaches 6, f(x) equals 36.

Visual Description: Graph of the quadratic function f(x) = x^2 on a set of Cartesian coordinate axes with vertex at the origin (0,0). A point is marked on the right branch of the parabola, with dashed lines extending down to x = 6 on the horizontal axis and left to y = 36 on the vertical axis, illustrating that as x approaches 6, f(x) equals 36.

\(f(x) = mx^2 + bx + c\) is quadratic function

graph is a parabola

\[ \lim_{x \to a} f(x) = f(a) = ma^2 + ba + c \]

Finding limits of polynomial functions

e.g.:

\( P(x) = 7x - 5 \)
\( P(x) = x^2 \)
\( P(x) = 9x^2 - 5x + 3 \)
\( P(x) = -9x^3 + 6x - 8 \)

In General:

\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_2 x^2 + a_1 x + a_0 \]

is a polynomial of degree \( n \).

\[ \lim_{x \to a} P(x) = P(a) \]

Cartesian coordinate graph of a polynomial function curve P(x) drawn in blue across the x and y axes. A point on the curve is marked with a black dot corresponding to x = a on the horizontal axis and P(a) on the vertical axis, connected by dashed projection lines to illustrate that the limit of P(x) as x approaches a is equal to P(a).

Visual Description: Cartesian coordinate graph of a polynomial function curve P(x) drawn in blue across the x and y axes. A point on the curve is marked with a black dot corresponding to x = a on the horizontal axis and P(a) on the vertical axis, connected by dashed projection lines to illustrate that the limit of P(x) as x approaches a is equal to P(a).

e.g.:

\[ \begin{aligned} \lim_{x \to 2} x^3 - 9x + 5 &= 2^3 - 9(2) + 5 \\ &= 8 - 18 + 5 \\ &= -5 \end{aligned} \]

\[ \lim_{x \to 2} x^3 - 9x + 5 = -5 \]

\[ \lim_{x \to 2} 5 = 5 \]

\[ \lim_{x \to 2} x = 2 \]

\[ \lim_{x \to 2} 9x = 18 = 9 \cdot \lim_{x \to 2} x = 9 \cdot 2 \]

\[ \lim_{x \to 2} x^3 = 2^3 = 8 = \left(\lim_{x \to 2} x\right)^3 \]

\[ \lim_{x \to 2} x^3 - 9x + 5 = \lim_{x \to 2} x^3 - \lim_{x \to 2} 9x + \lim_{x \to 2} 5 \]

Limit Laws: \( \lim_{x \to a} f(x) = L_1, \quad \lim_{x \to a} g(x) = L_2 \)

\[ \lim_{x \to a} (f(x) + g(x)) = L_1 + L_2 \]
\[ \lim_{x \to a} (f(x) - g(x)) = L_1 - L_2 \]
\[ \lim_{x \to a} (f(x) g(x)) = L_1 L_2 \]
\[ \lim_{x \to a} (f(x))^k = (L_1)^k \]
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L_1}{L_2} \]
only when \( L_2 \neq 0 \)
\[ \lim_{x \to a} \sqrt[k]{f(x)} = \sqrt[k]{L_1} \]
Be careful when \( k \) is even.

1. \( \lim_{x \to 8} \frac{x^2 - 4x - 21}{x - 7} \)

plug in \( x = 8 \)

\[ \lim_{x \to 8} x^2 - 4x - 21 = 8^2 - 4(8) - 21 = 11 \] \[ \lim_{x \to 8} x - 7 = 8 - 7 = 1 \neq 0 \] \[ \lim_{x \to 8} \frac{x^2 - 4x - 21}{x - 7} = 11 \]

2. \( \lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 7} \)

CANNOT plug in because

\[ \lim_{x \to 7} x - 7 = 0 \]

in fact

\[ \lim_{x \to 7} x^2 - 4x - 21 = 0 \]

if you plug in you get \( \frac{0}{0} \) indeterminate form

Finding limits of Rational Functions

\[ R(x) = \frac{P(x)}{Q(x)}, \quad P, Q \text{ are polynomials.} \]

If \( \lim_{x \to a} Q(x) = Q(a) \neq 0 \), then

\[ \lim_{x \to a} R(x) = \lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)} = R(a) \]

CAN PLUG IN
If \( \lim_{x \to a} Q(x) = Q(a) = 0 \)

CANNOT PLUG IN

Example:

\[ \lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 7} \] \[ \frac{7^2 - 4(7) - 21}{7 - 7} = \frac{0}{0} \]

factorize Numerator: \( x^2 - 4x - 21 = (x-7)(x+3) \)

\( \lim_{x \to 7} f(x) \) means evaluate \( f(x) \) when \( x \) is close to \( 7 \), not \( = 7 \)

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

CAN plug in \( x=7 \)

\[ = \underline{10} \]

Eg: \( \lim_{x \to 0} \frac{\sqrt{x+25}-5}{x} \)

if you plug in we get \( \frac{0}{0} \)

if you see a Radical, Try to Rationalize

\[ \frac{\sqrt{x+25}-5}{x} = \frac{(\sqrt{x+25}-5)(\sqrt{x+25}+5)}{x(\sqrt{x+25}+5)} = \frac{x+25-25}{x(\sqrt{x+25}+5)} \] \[ = \frac{x}{x(\sqrt{x+25}+5)} \] \[ = \frac{1}{\sqrt{x+25}+5} \] \[ \lim_{x \to 0} \frac{\sqrt{x+25}-5}{x} = \lim_{x \to 0} \frac{1}{\sqrt{x+25}+5} \]

CAN Plug in

\[ = \frac{1}{10} \]

Example: Evaluate \( \lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3} \)

Get \( \frac{0}{0} \) if you plug in \( x = 3 \)

Numerator = difference of two fractions \( \left. \vphantom{\text{difference of two fractions}} \right\} \) Make it one fraction

\[ \frac{1}{x} - \frac{1}{3} = \frac{3-x}{3x} \] \[ \lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3} = \lim_{x \to 3} \frac{\frac{3-x}{3x}}{x - 3} = \lim_{x \to 3} \frac{3-x}{3x(x-3)} \] \[ = \lim_{x \to 3} \frac{-1}{3x} \]

CAN plug in

\[ = -\frac{1}{9}. \]

Squeeze/ Sandwich Theorem

Visual Description: Hand-drawn Cartesian coordinate graph illustrating the Squeeze/Sandwich Theorem. Three continuous functions of x are plotted: the upper curve h(x) in purple, the lower curve g(x) in blue, and the middle curve f(x) in green trapped between them. Specific points on each curve at x = a are marked with dots: a purple point on h(x), a green point on f(x), and a blue point on g(x), demonstrating the inequality g(x) ≤ f(x) ≤ h(x).

Suppose

\[ g(x) \le f(x) \le h(x) \]

Then

\[ \lim_{x \to a} g(x) \le \lim_{x \to a} f(x) \le \lim_{x \to a} h(x) \]

if \( f(x) \) is sandwiched b/w \( g(x) \) & \( h(x) \), then \( \lim_{x \to a} f(x) \) is also sandwiched b/w the corresponding limits

e.g. Evaluate \(\lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right)\)

CANNOT plug in \(x = 0\)

Recall: \(-1 \le \cos\theta \le 1\) for any \(\theta\)

\(-1 \le \cos\left(\frac{2}{x}\right) \le 1\) for any \(x \neq 0\)

multiply \(x^4\)

\[ -x^4 \le x^4 \cos\left(\frac{2}{x}\right) \le x^4 \]

Cartesian coordinate graph illustrating the Sandwich (Squeeze) Theorem. A purple upward-opening curve labeled x^4 serves as the upper bound, and a blue downward-opening curve labeled -x^4 serves as the lower bound, both meeting at the origin (0,0). An oscillating green curve labeled x^4 cos(2/x) is trapped entirely between the two bounds, showing that its oscillations are squeezed to 0 as x approaches 0.

Visual Description: Cartesian coordinate graph illustrating the Sandwich (Squeeze) Theorem. A purple upward-opening curve labeled x^4 serves as the upper bound, and a blue downward-opening curve labeled -x^4 serves as the lower bound, both meeting at the origin (0,0). An oscillating green curve labeled x^4 cos(2/x) is trapped entirely between the two bounds, showing that its oscillations are squeezed to 0 as x approaches 0.

Applying Sandwich theorem

\[ \lim_{x \to 0} (-x^4) \le \lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right) \le \lim_{x \to 0} x^4 \] \[ 0 \le \lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right) \le 0 \]

therefore

\[ \lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right) = 0 \]