\(\lim_{x \to 3} 2x - 4 = 3\)

Suppose \(\epsilon > 0\). Let \(\delta = \dfrac{\epsilon}{2} > 0\)

Suppose \(0 < |x-3| < \delta\)

\(|2x - 4 - 2|\) = |2||x - 3|

From \(0 < |x-3| < \delta\), \(|2||x - 3| < 2\delta\)

\(|2||x - 3| < 2 \dfrac{\epsilon}{2}\)

\(|2||x - 3| < \epsilon\)

Hence, \(f(x) - L < \epsilon\)

\(\lim_{t \to 1} 5 - 3t = 2\)

Suppose \(\epsilon > 0\). Let \(\delta = \dfrac{\epsilon}{3} > 0\)

Suppose \(0 < |t-1| < \delta\)

\(|5 - 3t - 2| = |3||1 - t|\)

From \(0 < |t-1| < \delta\), \(|3||t - 1| < 3\delta\)

\(|3||t - 1| < \epsilon\)

Hence, \(f(x) - L < \epsilon\)

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

Suppose \(\epsilon > 0\). Let \(\delta = \epsilon > 0\)

Suppose \(0 < |x-(-2)| < \delta\)

\(|\dfrac{x^2 - 4}{x + 2} + 4| = |x + 2| = |x - (-2)| < \delta\)

Hence, \(f(x) - L < \epsilon\)

\(\lim_{x \to 2} \dfrac{x^2 - x - 2}{x - 2} = 3\)

Suppose \(\epsilon > 0\). Let \(\delta = \epsilon > 0\)

Suppose \(0 < |x-2| < \delta\)

Since \(|x - 2| > 0\), \(x \neq 2\) and hence the function is defined.

\(| \dfrac{x^2 - x - 2}{x - 2} - 3| = |x - 2|\)

We know that \(|x - 2| < \delta\), So, $|x - 2| < ε $

\(\lim_{r \to 3} \dfrac{3r^2 - 8r - 3}{2r - 6} = 5\)

Suppose \(\epsilon > 0\). Let \(\delta = \dfrac{2}{3}\epsilon > 0\)

Suppose \(0 < |r-3| < \delta\)

Since \(0 < |r-3|\), \(r \neq 3\) and hence the function is defined.

\(|\dfrac{3r^2 - 8r - 3}{2r - 6} - 5| = |\dfrac{3}{2}(r - 3)|\)

We know that \(| r - 3 | < \delta\)

\(|\dfrac{3}{2}(r - 3) | < \dfrac{3}{2}\delta < \epsilon\)

\(\lim_{x \to 2} \dfrac{x^2 + 2x - 8}{3x - 6} = 2\)

Suppose \(\epsilon > 0\). Let \(\delta = 2\epsilon > 0\)

Suppose \(0 < |x-2| < \delta\)

Since \(0 < |x-2|\), \(x \neq 2\) and hence the function is defined.

\(|\dfrac{x^2 + 2x - 8}{3x - 6} - 2| = \dfrac{1}{2}(x - 2)\)

We know that \(|x - 2| < \sigma\)

\(\dfrac{1}{2}(x - 2) < 2.\dfrac{\epsilon}{2}\)

\(\dfrac{1}{2}(x - 2) < \epsilon\)

\(\lim_{u \to -1} \dfrac{u^2 + 4u + 3}{u + 1} + u = 1\)

Suppose \(\epsilon > 0\). Let \(\delta = \dfrac{\epsilon}{2} > 0\)

Suppose \(0 < |u-(-1)| < \delta\)

Since \(|u-(-1)| > 0\), \(u \neq -1\) and hence the function is defined.

\(\dfrac{u^2 + 4u + 3}{u + 1} + u - 1| = |2||u+1|\)

We know that \(|u - (-1)| < \delta\)

\(2 |u + 1| < 2\delta\)

\(2 |u + 1| < \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \dfrac{\epsilon}{2})\)

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \dfrac{\epsilon}{2}\).

Now suppose \(0 < |x-2| < \delta\). Then,

\(|x - 2| < \delta \leq 1\). So, \(0 < x - 1 < 2\).

Thus, \(|x^2 - 3x + 5 - 3| = |(x - 1)(x-2)| < 2|(x - 2)| < 2\delta < 2\dfrac{\epsilon}{2} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(\dfrac{1}{2}, \dfrac{\epsilon}{12})\)

Notice that we have \(\delta > 0, \delta \leq \dfrac{1}{2} and \delta \leq \dfrac{\epsilon}{12}\).

Now suppose \(0 < |v-1| < \delta\). Then,

\(|v - 1| < \delta \leq \dfrac{1}{2}\). So, \(1 < 2v < 3\).

From \(1 < 2v\), we infer \(\dfrac{1}{3v - 1} < 2\).

Thus, \(|\dfrac{4}{3v - 1} - 2| = |\dfrac{6(v-1)}{1 - 3v}| = |\dfrac{6(v-1)}{3v - 1}| < 6(v-1)*2 <= 12\sigma <= 12\dfrac{\epsilon}{2} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \epsilon)\)

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \epsilon\).

Now suppose \(0 < |t-1| < \delta\). Then,

\(|t - 1| < \delta \leq 1\). So, \(0 < t < 2\).

From \(0 < t\), we know that \(\dfrac{1}{t + 1} < 1\)

Thus, \(|\dfrac{2 - 2t}{t^2 - 1} + 1| = |\dfrac{t - 1)}{t + 1}| < |t-1| < \delta \leq \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = 1\)

Now suppose \(0 < |x -3| < 1\). Solving it, we get \(1 < \dfrac{x}{2} < 2\)

Since \(\lfloor \dfrac{x}{2} \rfloor\) is always an integer from its range of \(1 < \dfrac{x}{2} < 2\), we can determinte that \(\lfloor \dfrac{x}{2} \rfloor\) will be \(1\).

So, \(|\lfloor \dfrac{x}{2} \rfloor - 1| = 0 < \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(2, \dfrac{\epsilon}{28})\).

Notice that we have \(\delta > 0, \delta \leq 2 and \delta \leq \dfrac{\epsilon}{28}\).

Now suppose \(0 < |x-2| < \delta\). Then,

\(|x - 2| < \delta \leq 2\). So, \(x^2 < 16\) and \(2x < 8\).

From \(x^2 < 16\) and \(2x < 8\), \(x^2 + 2x < 24\). So, \(x^2 + 2x + 4< 28\)

Thus, \(|x^3 - 8| = |(x-2)(x^2 + 4 + 2x)| < |(x - 2)|28 < ||(x-2).28| < 28.\delta <= 28\dfrac{\epsilon}{28} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \dfrac{3\epsilon}{2})\).

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \dfrac{3\epsilon}{2}\).

Now suppose \(0 < |w-2| < \delta\). Then,

\(|w-2| < \delta \leq 1\). So, From \(1 < w\), we can infer that \(\dfrac{1}{\sqrt{2w + 5} + 3} < \dfrac{1}{\sqrt{7} + 3}\)

We know that \(\dfrac{1}{\sqrt{7} + 3} < \dfrac{1}{3}\)

So by using transitive property,

\(\dfrac{1}{\sqrt{2w + 5} + 3} < \dfrac{1}{3}\)

\(|\sqrt{2w + 5} - 3| = |\dfrac{2(w - 2)}{\sqrt{2w + 5} + 3}|\)

( Note that the trick in the above reduction is to multiply by \(\dfrac{{\sqrt{2w + 5} + 3}}{{\sqrt{2w + 5} + 3}}\))

\(|\dfrac{2(w - 2)}{\sqrt{2w + 5} + 3}| < \dfrac{2(w-2)}{3} < \dfrac{2\delta}{3} \leq \dfrac{2.3\delta}{3.2} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(2, \dfrac{\epsilon}{8})\).

Notice that we have \(\delta > 0, \delta \leq 2 and \delta \leq \dfrac{\epsilon}{8}\).

Now suppose \(0 < |t-3| < \delta\). Then,

\(|t^2 - 9| = |(t+3)(t-3)|\)

Since \(|t-3| < \delta \leq 2\), we have \(1 < t < 5\). Thus, \(4 < t + 3 < 8\)

\((t+3)(t-3) = |t-3|(t+3) < 8|t-3| < 8\delta \leq \dfrac{8\epsilon}{8} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \dfrac{\epsilon}{4})\).

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \dfrac{\epsilon}{4}\).

Now suppose \(0 < |x-1| < \delta\). Then, \(0 < x < 2\). So, \(x + 2 < 4\).

\(|x^2 + x - 2| = |(x + 2)(x - 1)| < 4|x-1| < 4\delta \leq \dfrac{4\epsilon}{4} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \dfrac{\epsilon}{9})\).

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \dfrac{\epsilon}{9}\).

Now suppose \(0 < |x-2| < \delta\). Then, \(1 < x < 3\). So, \(3x < 9\).

\(|3x^2 - 6x - 0| = |3x(x-2)| < |x-2|9 < 9\delta <= \dfrac{9\epsilon}{9} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(3, \epsilon)\).

Notice that we have \(\delta > 0, \delta \leq 3 and \delta \leq \epsilon\).

Now suppose \(0 < |x-(-1)| < \delta\). Then, \(\delta \leq 3\). So, \(|x-(-1)| < 3\) and therefore \(x-1<1\).

\(|\dfrac{x^3 + x^2 -3x - 3}{x+1} + 2| = |x^2 - 1| = |(x-1)(x+1)| < |x+1| < |x - (-1)| < \delta = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \dfrac{\epsilon}{2})\).

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \dfrac{\epsilon}{2}\).

Now suppose \(0 < |u-3| < \delta\). Then since, \(\delta \leq 1\). we have \(2 < u <4\) and therefore \(\dfrac{1}{u-1} < 1\).

$|\dfrac{4}{u-1} - 2| = $\dfrac{2(u-3)}{u-1}| < |2(u-3)| < 2δ < \dfrac{2\epsilon}{2} = \*

Suppose \(\epsilon > 0\). Let \(\delta = min(4, \dfrac{\epsilon}{2})\).

Notice that we have \(\delta > 0, \delta \leq 4 and \delta \leq \dfrac{\epsilon}{2}\).

Now suppose \(0 < |t-5| < \delta\). Then since, \(\delta \leq 4\). we have \(1 < t <9\) and therefore \(\dfrac{1}{t} < 1\).

\(|\dfrac{3t - 10}{t} - 1| = |\dfrac{2(t-5)}{t}| < 2(t-5) < 2\delta \leq \dfrac{2\epsilon}{2} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(3, \sqrt{\dfrac{15\epsilon}{2}})\).

Notice that we have \(\delta > 0, \delta \leq 2 and \delta \leq \dfrac{15\epsilon}{2}\).

Now suppose \(0 < |z-1| < \delta\). Then since, \(\delta \leq 3\). we have \(-3 < z-1 <3\) and therefore \(9 < (z-1)^2, 1 < (z + 1)^2 and -7 < z - 5\)

\(|\dfrac{z-1}{2z^2 + z - 3} - \dfrac{1}{5}| = |\dfrac{-2(z-1)^2}{5((z+1)^2 + (z-1)^2 + (z - 5))}|\)

We know that, \(\dfrac{1}{(z-1)^2+(z+1)^2 + (z-5)} < \dfrac{1}{3}\)

\(|\dfrac{-2(z-1)^2}{5((z+1)^2 + (z-1)^2 + (z - 5))}| < \dfrac{-2(z-1)^2}{5}\dfrac{1}{3} < \dfrac{2\delta^2}{16} \leq \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \dfrac{\epsilon}{4})\).

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \dfrac{\epsilon}{4}\).

Now suppose \(0 < |y-1| < \delta\). Then since, \(\delta \leq 1\). we have \(0 < y <2\) and therefore \(y^2 < 4\)

\(|y^3 - y^2 + 7 = 7| = |y^2(y-1)| < 4|y-1| < 4\delta \leq 4\dfrac{\epsilon}{4} = \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \sqrt{\dfrac{\epsilon}{2}})\).

Notice that we have \(\delta > 0, \delta \leq 1 and \delta \leq \sqrt{\dfrac{\epsilon}{2}}\).

Now suppose \(0 < |y-0| < \delta\). Then since, \(\delta \leq 1\). we have \(-1 < y <1\) and therefore \(1-y < 2\)

\(|y^3 - y^2 + 7 - 7| = |y^2(1-y)| < 2y^2 < 2\delta^2 \leq \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, 2(3 + \sqrt[2]{3})\epsilon)\).

Notice that we have \(\delta > 0, \delta \leq 1\) and \(\delta \leq 2(3 + \sqrt[2]{3})\epsilon\).

Suppose \(0 < |x-4| < \epsilon\). Then since, \(\delta \leq 1\), we have \(3 < x < 5\) and therefore \(\dfrac{1}{x + 2\sqrt{x} < \dfrac{1}{3 + 2\sqrt{3}}}\)

\(|\dfrac{1}{\sqrt{x}} - \dfrac{1}{2}| = \dfrac{-1(4-x)}{2(x + 2\sqrt{x})} < \dfrac{|-1(4-x)|}{2(3+2\sqrt{3})} < \dfrac{|4-x|}{2(x+2\sqrt{3})} < \dfrac{\epsilon}{2(3 + 2\sqrt{3})} \leq \epsilon\)

Suppose \(\epsilon > 0\). Let \(\delta = min(1, \sqrt{\epsilon})\).

Notice that we have \(\delta > 0, \delta \leq 1\) and \(\delta \leq \sqrt{\epsilon}\).

Suppose \(0 < |x-0| < \epsilon\). Then since, \(\delta \leq 1\), we have \(-1 < x < 1\) and therefore \(\dfrac{1}{\sqrt{x^2 + 1}-1} < \dfrac{1}{\sqrt{2}-1} < 1\)

\(|\sqrt{x^2 + 1} - 1| = |\dfrac{x^2}{\sqrt{x^2 + 1} + 1}| < |x^2| < \delta^2 \leq \epsilon\)