The problem asks to calculate the limits of various functions, indicating when they do not exist and why. It also asks to discuss the continuity of each function at the specified point, illustrate the epsilon-delta definition of a limit with a specific example, and prove statements using the epsilon-delta definition of the limit.

1. Problem Description

The problem asks to calculate the limits of various functions, indicating when they do not exist and why. It also asks to discuss the continuity of each function at the specified point, illustrate the epsilon-delta definition of a limit with a specific example, and prove statements using the epsilon-delta definition of the limit.

2. Solution Steps

(a) $\lim_{x\to 5} \frac{x^2 - 6x + 5}{x - 5}$

First, factor the numerator: $x^2 - 6x + 5 = (x - 5)(x - 1)$. Then,

$\lim_{x\to 5} \frac{(x - 5)(x - 1)}{x - 5} = \lim_{x\to 5} (x - 1) = 5 - 1 = 4$

(b) $\lim_{t\to -3} \frac{t^2 - 9}{2t^2 + 7t + 3}$

Factor the numerator and denominator:

$t^2 - 9 = (t - 3)(t + 3)$

$2t^2 + 7t + 3 = (2t + 1)(t + 3)$

Then,

$\lim_{t\to -3} \frac{(t - 3)(t + 3)}{(2t + 1)(t + 3)} = \lim_{t\to -3} \frac{t - 3}{2t + 1} = \frac{-3 - 3}{2(-3) + 1} = \frac{-6}{-6 + 1} = \frac{-6}{-5} = \frac{6}{5}$

(c) $\lim_{x\to -2} \frac{x + 2}{x^3 + 8}$

Factor the denominator using the sum of cubes formula:

$x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

Then,

$\lim_{x\to -2} \frac{x + 2}{(x + 2)(x^2 - 2x + 4)} = \lim_{x\to -2} \frac{1}{x^2 - 2x + 4} = \frac{1}{(-2)^2 - 2(-2) + 4} = \frac{1}{4 + 4 + 4} = \frac{1}{12}$

(d) $\lim_{h\to 0} \frac{(3 + h)^{-1} - 3^{-1}}{h}$

Rewrite the expression:

$\frac{\frac{1}{3 + h} - \frac{1}{3}}{h} = \frac{\frac{3 - (3 + h)}{3(3 + h)}}{h} = \frac{\frac{-h}{3(3 + h)}}{h} = \frac{-h}{3h(3 + h)} = \frac{-1}{3(3 + h)}$

Then,

$\lim_{h\to 0} \frac{-1}{3(3 + h)} = \frac{-1}{3(3 + 0)} = \frac{-1}{9}$

(e) $\lim_{x\to -1} \frac{x^4 - 1}{x + 1}$ if $x \ne -1$, and 0 if $x = -1$

Factor the numerator:

$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$

Then, for $x \ne -1$,

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

Since the limit as $x$ approaches $-1$ is $-4$, but the function is defined as $0$ when $x = -1$, the limit exists and equals $-4$.

(f) $\lim_{x\to \frac{1}{2}} \frac{2x - 1}{|2x^3 - x^2|}$

Factor the denominator:

$|2x^3 - x^2| = |x^2(2x - 1)| = x^2|2x - 1|$

Then,

$\lim_{x\to \frac{1}{2}} \frac{2x - 1}{x^2|2x - 1|}$

Consider the limit from the left ($x \to \frac{1}{2}^-$), then $2x - 1 < 0$, so $|2x-1| = -(2x-1) = 1 - 2x$.

$\lim_{x\to \frac{1}{2}^-} \frac{2x - 1}{x^2(1 - 2x)} = \lim_{x\to \frac{1}{2}^-} \frac{-(1 - 2x)}{x^2(1 - 2x)} = \lim_{x\to \frac{1}{2}^-} \frac{-1}{x^2} = \frac{-1}{(\frac{1}{2})^2} = \frac{-1}{\frac{1}{4}} = -4$

Consider the limit from the right ($x \to \frac{1}{2}^+$), then $2x - 1 > 0$, so $|2x-1| = 2x-1$.

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

Since the left-hand limit and the right-hand limit are not equal, the limit does not exist.

(g) $\lim_{x\to -2} [x] + [-x]$ where $[x]$ is the greatest integer function.

If $x$ is not an integer, then $[x] + [-x] = -1$.

If $x$ is an integer, then $[x] + [-x] = x - x = 0$

$\lim_{x \to -2} [x] + [-x] = 0$. However, as $x$ approaches -2 from the left or right, it will not be exactly -2, thus the value will be -

1. Therefore, the limit is -

1.

(h) $\lim_{t\to 1} H(t)$ where $H(t) = \begin{cases} 0 \text{ if } t < 0 \\ 1 \text{ if } t \ge 0 \end{cases}$

Since $1 > 0$, as $t$ approaches $1$, we have $t > 0$, so $H(t) = 1$.

$\lim_{t\to 1} H(t) = 1$

II. Continuity discussion:

(a) The function $f(x) = \frac{x^2-6x+5}{x-5}$ is not continuous at $x=5$ because it is not defined there.

(b) The function $f(t) = \frac{t^2-9}{2t^2+7t+3}$ is not continuous at $t=-3$ and $t=-\frac{1}{2}$ because it is not defined there.

(c) The function $f(x) = \frac{x+2}{x^3+8}$ is not continuous at $x=-2$ because it is not defined there.

(d) The function $f(h) = \frac{(3+h)^{-1}-3^{-1}}{h}$ is not continuous at $h=0$ because it is not defined there.

(e) The function is defined as $f(x) = \frac{x^4-1}{x+1}$ for $x\ne -1$ and $f(x) = 0$ for $x=-1$. Since $\lim_{x\to -1} f(x) = -4 \ne f(-1) = 0$, the function is not continuous at $x=-1$.

(f) The function $f(x) = \frac{2x-1}{|2x^3-x^2|}$ is not continuous at $x=1/2$ because it is not defined there.

(g) The function $f(x) = [x] + [-x]$ is not continuous at $x = -2$.

(h) The function $H(t)$ is continuous everywhere except at $t = 0$. Since $t \to 1$, the function is continuous at $t=1$.

III. Illustrate the precise definition of limits by finding the values of $\delta$ that correspond to the given values of $\epsilon$:

$\lim_{x\to -2} \frac{1}{2}x + 3 = 2$. $\epsilon = 1, \epsilon = 0.01$

$|\frac{1}{2}x + 3 - 2| < \epsilon$

$|\frac{1}{2}x + 1| < \epsilon$

$|\frac{1}{2}(x + 2)| < \epsilon$

$\frac{1}{2}|x + 2| < \epsilon$

$|x + 2| < 2\epsilon$

Thus $\delta = 2\epsilon$

If $\epsilon = 1$, $\delta = 2(1) = 2$.

If $\epsilon = 0.01$, $\delta = 2(0.01) = 0.02$.

IV. Prove the statement using the $\epsilon$, $\delta$ definition of the limit

(a) $\lim_{x\to -3} (1 - 4x) = 13$

$|1 - 4x - 13| < \epsilon$

$|-4x - 12| < \epsilon$

$|-4(x + 3)| < \epsilon$

$4|x + 3| < \epsilon$

$|x + 3| < \frac{\epsilon}{4}$

Choose $\delta = \frac{\epsilon}{4}$.

(b) $\lim_{x\to -2} (3x + 5) = -1$

$|3x + 5 - (-1)| < \epsilon$

$|3x + 6| < \epsilon$

$|3(x + 2)| < \epsilon$

$3|x + 2| < \epsilon$

$|x + 2| < \frac{\epsilon}{3}$

Choose $\delta = \frac{\epsilon}{3}$.

(c) $\lim_{x\to -1.5} \frac{9 - 4x^2}{3 + 2x} = 6$

$\frac{9-4x^2}{3+2x} = \frac{(3-2x)(3+2x)}{3+2x} = 3-2x$, for $x \ne -1.5$

$|3 - 2x - 6| < \epsilon$

$|-2x - 3| < \epsilon$

$|-2(x + 1.5)| < \epsilon$

$2|x + 1.5| < \epsilon$

$|x + 1.5| < \frac{\epsilon}{2}$

Choose $\delta = \frac{\epsilon}{2}$.

(d) $\lim_{x\to a} x = a$

$|x - a| < \epsilon$

Choose $\delta = \epsilon$.

3. Final Answer

(a) 4

(b) 6/5

(c) 1/12

(d) -1/9

(e) -4

(f) Does not exist

(g) -1

(h) 1

II. Continuity discussion: see step

2.

III. $\delta = 2$ for $\epsilon = 1$; $\delta = 0.02$ for $\epsilon = 0.01$

IV. (a) $\delta = \frac{\epsilon}{4}$

(b) $\delta = \frac{\epsilon}{3}$

(c) $\delta = \frac{\epsilon}{2}$

(d) $\delta = \epsilon$