Session 02 - Introduction to Limits

1. Left-Hand Trend ($x \to 3^-$): Plug values slightly less than $3$ into the simplified path rule $f(x) = x + 3$:
   • $f(2.9) = 2.9 + 3 = 5.9$
   • $f(2.99) = 2.99 + 3 = 5.99$
   • $f(2.999) = 2.999 + 3 = 5.999$
2. Right-Hand Trend ($x \to 3^+$): Plug values slightly greater than $3$ into the path rule:
   • $f(3.1) = 3.1 + 3 = 6.1$
   • $f(3.01) = 3.01 + 3 = 6.01$
   • $f(3.001) = 3.001 + 3 = 6.001$
3. Analyze Convergence: Assemble the table:

As $x$ pinches closer to $3$ from both sides, the output values head toward the distinct value $6$. Therefore, the limit is:

1. Determine the Function Value $f(-1)$: Look precisely along the vertical grid line $x = -1$. There is an open circle at a height of $2$ and a solid dot at a height of $1$. The solid dot represents the actual function value, so $f(-1) = 1$.
2. Determine the Limit $\lim_{x \to -1} f(x)$: Trace along the lines of the graph as you approach $x = -1$ from both the left and right. Both paths lead directly to the open circle at height $y = 2$, so $\lim_{x \to -1} f(x) = 2$.
3. Compare Results: Note that $\lim_{x \to -1} f(x) \neq f(-1)$ since $2 \neq 1$. This proves that a limit depends entirely on the expected trajectory of the path, independent of the actual point plotted at the destination.

• (a) According to the Conceptual Framework section, algebra deals with exact coordinates at $x=c$, whereas limits describe the trend of outputs as we approach the input $c$ from neighboring values.
• (b) In Worked Example B, tracing the graph from both sides of $x = -1$ leads to the open circle at $y = 2$, which gives the limit. The solid dot at $(-1, 1)$ represents the isolated function value, showing that path behavior (limit) can differ from the exact coordinate.
• (c) As outlined in the Engineering Applications section under "Signal Cutoff Gain", limits allow electrical engineers to compute signal stability and gain near mathematical singularities (like division by zero at boundary cutoffs) by examining approaching trends.

1. Simplify the Function: Factor the numerator:
   $$f(x) = \frac{(x-2)(x+2)}{x-2} = x+2 \quad (\text{for } x \neq 2)$$
2. Evaluate the Table:
   • $f(1.9) = 1.9+2 = 3.9$
   • $f(1.99) = 1.99+2 = 3.99$
   • $f(1.999) = 1.999+2 = 3.999$
   • $f(2.001) = 2.001+2 = 4.001$
   • $f(2.01) = 2.01+2 = 4.01$
   • $f(2.1) = 2.1+2 = 4.1$
3. Evaluate the Limits:
   • (a) As $x \to 2^-$, $f(x)$ approaches $4$.
   • (b) As $x \to 2^+$, $f(x)$ approaches $4$.
   • (c) Since the left-hand and right-hand limits match, the two-sided limit exists and is equal to $4$.
   • (d) Direct substitution of $x = 2$ gives $f(2) = \frac{0}{0}$, which is undefined. The limit describes what height the function approaches as we get closer to $x=2$ from the sides, which is $4$.

1. Part (a) - $f(-1)$: At $x = -1$, there is an open circle at $y = 2$ and a solid dot at $y = 1$. The solid dot represents the function value, so $f(-1) = 1$.
2. Part (b) - $\lim_{x \to -1} f(x)$: Approaching $x = -1$ from both sides, the curve targets the open circle at height $y = 2$. Thus, the limit is $2$.
3. Part (c) - $f(4)$: Looking at $x = 4$, there is a solid dot at $y = -2$ and an open circle at $y = -3$. Therefore, $f(4) = -2$.
4. Part (d) - $\lim_{x \to 4^+} f(x)$: Tracing the curve from the right toward $x = 4$, the curve approaches the solid dot at height $y = -2$.
5. Part (e) - $\lim_{x \to 4^-} f(x)$: Tracing the line from the left toward $x = 4$, the line leads to the open circle at height $y = -3$.
6. Part (f) - $\lim_{x \to 4} f(x)$: Since the left-hand limit ($-3$) and right-hand limit ($-2$) do not match, the two-sided limit does not exist ($\text{DNE}$).
7. Part (g) - $\lim_{x \to -4^-} f(x)$: Tracing from the left toward $x = -4$, the curve ends at the solid dot at height $y = 3$.
8. Part (h) - $\lim_{x \to -4} f(x)$: Tracing from the right toward $x = -4$, the curve approaches the open circle at $y = 4$. Since the left-hand limit ($3$) and right-hand limit ($4$) do not match, the two-sided limit is $\text{DNE}$.
9. Part (i) - $\lim_{x \to -5} f(x)$: The curve is continuous and smooth at $x = -5$, passing exactly through height $y = 2$ from both directions, so the limit is $2$.

1. Evaluate Left-Hand Limit: Trace the graph from the left toward $x = 0$ ($x < 0$). The function forms a flat line at a height of $-1$:
   $$\lim_{x \to 0^-} \frac{|x|}{x} = -1$$
2. Evaluate Right-Hand Limit: Trace the graph from the right toward $x = 0$ ($x > 0$). The function forms a flat line at a height of $1$:
   $$\lim_{x \to 0^+} \frac{|x|}{x} = 1$$
3. Determine Two-Sided Limit: Because the left-hand and right-hand limits do not match ($-1 \neq 1$), by the Existence Theorem, the two-sided limit does not exist:
   $$\lim_{x \to 0} \frac{|x|}{x} = \text{DNE}$$
   (Classification: Jump Discontinuity)

1. Numerical Evaluation: Pick values that approach $0$ from both sides:
   • $f(\pm 0.1) = \frac{1}{(0.1)^2} = 100$
   • $f(\pm 0.01) = \frac{1}{(0.01)^2} = 10,000$
   • $f(\pm 0.001) = \frac{1}{(0.001)^2} = 1,000,000$
2. Analyze Trend: As inputs get closer to $0$, the output values grow without bound toward positive infinity.
3. Conclude: Graphically, both sides of the curve shoot straight up along the vertical asymptote $x = 0$. Since the values do not approach a single real number, the limit does not exist:
   $$\lim_{x \to 0} \frac{1}{x^2} = \text{DNE} \quad (\text{or } +\infty) \quad \text{(Reason: Unbounded Behavior)}$$

1. Analyze Data points as $x \to 0^+$:
   • At $x = 0.01 \implies f(x) = \sin(100) \approx -0.5064$
   • At $x = 0.001 \implies f(x) = \sin(1000) \approx 0.8269$
   • At $x = 0.0001 \implies f(x) = \sin(10000) \approx -0.3056$
   • At $x = 0.00001 \implies f(x) = \sin(100000) \approx 0.0357$
2. Analyze Trend: As $x$ approaches $0$, the function value fluctuates rapidly and infinitely between $-1$ and $1$. It never settles down or stabilizes near a single number.
3. Conclude:
   $$\lim_{x \to 0} \sin\left(\frac{1}{x}\right) = \text{DNE} \quad \text{(Reason: Wildly Oscillating Behavior)}$$

• (a) The Conceptual Framework section lists three reasons a limit fails to exist:
  1. Jump Discontinuity: One-sided limits exist and are finite but differ.
  2. Unbounded Behavior: The function grows to positive or negative infinity (vertical asymptote).
  3. Wildly Oscillating Behavior: The function fluctuates endlessly between different values without converging.
• (b) In Worked Example A, tracing the graph of $f(x) = \frac{|x|}{x}$ near $x = 0$ yields a left-hand limit of $-1$ and a right-hand limit of $1$. Since the left and right trajectories approach different values, the two-sided limit does not exist (a Jump Discontinuity).
• (c) Tracing the data in Worked Example C shows that as $x \to 0$, the function values fluctuate back and forth between $-1$ and $1$ (e.g. going from $-0.5064 \to 0.8269 \to -0.3056$). The function never stabilizes or converges to a single coordinate, which is classified as Wildly Oscillating Behavior.

• Left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x - 1) = 2(3) - 1 = 5$
• Right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (10 - x) = 10 - 3 = 7$
• Since $5 \neq 7$, the two-sided limit fails to exist. Because both one-sided limits are finite numbers that do not match, this failure is classified as a Jump Discontinuity.

• As $x \to 2^-$, the denominator $x - 2$ is negative and approaches $0$, so $\frac{3}{x - 2} \to -\infty$.
• As $x \to 2^+$, the denominator $x - 2$ is positive and approaches $0$, so $\frac{3}{x - 2} \to +\infty$.
• Because the function values grow infinitely large (in absolute value) and do not approach a finite real number as $x \to 2$, the limit is $\text{DNE}$ due to Unbounded Behavior (a vertical asymptote at $x = 2$).

• As $x \to 0^+$, the angle $\theta = \frac{\pi}{x}$ approaches $+\infty$.
• The cosine function $\cos(\theta)$ oscillates continuously between $-1$ and $1$. As $x$ gets closer to $0$, these oscillations happen infinitely fast. For example:
  • $x = 1 \implies \cos(\pi) = -1$
  • $x = 0.5 \implies \cos(2\pi) = 1$
  • $x = 0.25 \implies \cos(4\pi) = 1$
  • $x = 2/3 \implies \cos(1.5\pi) = 0$
• The function values do not stabilize or settle near any single number. Thus, the limit is $\text{DNE}$ due to Wildly Oscillating Behavior.

1. Evaluate Function at Endpoints:
   • $f(0) = \sin(0) = 0$
   • $f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = 0.5 = \frac{1}{2}$
2. Apply Formula:
   $$\text{Average Rate} = \frac{f(\pi/6) - f(0)}{\pi/6 - 0} = \frac{0.5 - 0}{\pi/6} = \frac{0.5}{\pi/6} = \frac{3}{\pi} \approx 0.9549$$
   The average rate of change is $\frac{3}{\pi} \approx 0.9549$.

1. Find $s(1)$:
   $$s(1) = -16(1)^2 + 85(1) = 69\text{ ft}$$
2. Calculate Average Velocity over broad interval $[1, 2]$:
   • $s(2) = -16(2)^2 + 85(2) = 106\text{ ft}$
   • $\text{Average Velocity} = \frac{s(2) - s(1)}{2 - 1} = \frac{106 - 69}{1} = 37\text{ ft/sec}$
3. Shrink interval to $[1, 1.1]$:
   • $s(1.1) = -16(1.1)^2 + 85(1.1) = 74.14\text{ ft}$
   • $\text{Average Velocity} = \frac{74.14 - 69}{1.1 - 1} = \frac{5.14}{0.1} = 51.4\text{ ft/sec}$
4. Shrink interval to $[1, 1.01]$:
   • $s(1.01) = -16(1.01)^2 + 85(1.01) = 69.5284\text{ ft}$
   • $\text{Average Velocity} = \frac{69.5284 - 69}{1.01 - 1} = \frac{0.5284}{0.01} = 52.84\text{ ft/sec}$
5. Shrink interval to $[1, 1.001]$:
   • $s(1.001) = -16(1.001)^2 + 85(1.001) = 69.052984\text{ ft}$
   • $\text{Average Velocity} = \frac{69.052984 - 69}{1.001 - 1} = \frac{0.052984}{0.001} = 52.984\text{ ft/sec}$
6. Infer the Limit: As the time interval length approaches $0$, the average velocities ($37 \to 51.4 \to 52.84 \to 52.984$) converge toward $53\text{ ft/sec}$. This is the estimated instantaneous velocity at $t = 1$.

• (a) According to the Conceptual Framework section, the average rate of change corresponds to the slope of a secant line between two distinct points. The instantaneous rate of change corresponds to the slope of a tangent line at a single point.
• (b) As explained in the Conceptual Framework, substituting $h = 0$ directly into the rate expression yields $\frac{f(x_0) - f(x_0)}{0} = \frac{0}{0}$, which is undefined. We must use the limiting process ($h \to 0$) to evaluate the approaching behavior.
• (c) Tracing the solution in Worked Example B, the sequence of average velocities calculated over shrinking intervals is:
  • $[1, 2] \implies 37\text{ ft/sec}$
  • $[1, 1.1] \implies 51.4\text{ ft/sec}$
  • $[1, 1.01] \implies 52.84\text{ ft/sec}$
  • $[1, 1.001] \implies 52.984\text{ ft/sec}$ These values converge to $53\text{ ft/sec}$ as the time interval approaches 0.

1. Evaluate at $x = 2$:
   $$f(2) = 3(2)^2 - 3(2) + 5 = 12 - 6 + 5 = 11$$
2. Evaluate at $x = -1$:
   $$f(-1) = 3(-1)^2 - 3(-1) + 5 = 3 + 3 + 5 = 11$$
3. Compute the average rate of change:
   $$\text{Average Rate} = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{11 - 11}{3} = 0$$

1. Evaluate at $t = \frac{\pi}{6}$:
   $$g\left(\frac{\pi}{6}\right) = 3\left[\cos\left(\frac{\pi}{6}\right)\right]^2 = 3\left(\frac{\sqrt{3}}{2}\right)^2 = 3\left(\frac{3}{4}\right) = \frac{9}{4}$$
2. Evaluate at $t = 0$:
   $$g(0) = 3[\cos(0)]^2 = 3(1)^2 = 3$$
3. Compute the average rate of change:
   $$\text{Average Rate} = \frac{g(\pi/6) - g(0)}{\pi/6 - 0} = \frac{\frac{9}{4} - 3}{\frac{\pi}{6}} = \frac{-\frac{3}{4}}{\frac{\pi}{6}} = -\frac{3}{4} \cdot \frac{6}{\pi} = -\frac{18}{4\pi} = -\frac{9}{2\pi} \approx -1.4324$$

1. Find Initial Value:
   $$s(1) = 20(1) - 2(1)^2 = 18\text{ m}$$
2. Compute Average Velocities:
   • i. Interval $[1, 1.1]$:
     • $s(1.1) = 20(1.1) - 2(1.21) = 22 - 2.42 = 19.58\text{ m}$
     • $\text{Average Velocity} = \frac{19.58 - 18}{1.1 - 1} = \frac{1.58}{0.1} = 15.8\text{ m/s}$
   • ii. Interval $[1, 1.01]$:
     • $s(1.01) = 20(1.01) - 2(1.01)^2 = 20.2 - 2.0402 = 18.1598\text{ m}$
     • $\text{Average Velocity} = \frac{18.1598 - 18}{1.01 - 1} = \frac{0.1598}{0.01} = 15.98\text{ m/s}$
   • iii. Interval $[1, 1.001]$:
     • $s(1.001) = 20(1.001) - 2(1.001)^2 = 20.02 - 2.004002 = 18.015998\text{ m}$
     • $\text{Average Velocity} = \frac{18.015998 - 18}{1.001 - 1} = \frac{0.015998}{0.001} = 15.998\text{ m/s}$
   • iv. Interval $[1, 1.0001]$:
     • $s(1.0001) = 20(1.0001) - 2(1.0001)^2 = 20.002 - 2.00040002 = 18.00159998\text{ m}$
     • $\text{Average Velocity} = \frac{18.00159998 - 18}{1.0001 - 1} = \frac{0.00159998}{0.0001} = 15.9998\text{ m/s}$
3. Estimate Instantaneous Velocity: Looking at the sequence of average velocities:
   $$15.8 \to 15.98 \to 15.998 \to 15.9998$$
   As the time interval length approaches zero, the values converge to $16\text{ m/s}$.