\(\lim_{x \to \infty} \dfrac{\dfrac{1}{x} - 7}{x^3 - x^2}\)

\(= \lim_{x \to \infty} \dfrac{\dfrac{1}{x} - 7}{x^3(1 - \dfrac{1}{x})}\)

\(\lim_{x \to \infty} \dfrac{1}{x} - 7 = -7\)

\(\lim_{x \to \infty} 1 - \dfrac{1}{x} = 1 - 0 = 1\)

\(\lim_{x \to \infty} \dfrac{1}{x^3} = 0\)

Combining them, \(\dfrac{0.(-7)}{1} = 0\)

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

The numerator approaches 1 and the denominator approaches 0, so the limit is undefined. (From Theorem 2.4.8)

As \(x \to 3^-\), \(x - 3 \to 0^-\)

By part 2 of Theorem 2.6.2, as \(x \to 3^{\neq}\), \(\dfrac{1}{x-3} \to -\infty\)

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

\(\lim_{x \to 3^-} \dfrac{x-4}{x-3}\)

The numerator approaches -1 and the denominator approaches 0, so the limit is undefined. (From Theorem 2.4.8)

As \(x \to 3^-\), \(x - 3 \to 0^-\).

By part 2 of Theorem 2.6.2, as \(x \to 3^{\neq}\), \(\dfrac{1}{x-3} \to -\infty\)

Combining this with the fact as \(x \to 3^-\), \(x-4 \to -1\), we can use theorem 2.5.14,

\(\lim_{x \to 3^-} \dfrac{x-4}{x-3} = \infty\)

\(\lim_{t \to 0^+} \dfrac{t^2 + 3t}{t^3 - t^2}\)

\(= \lim_{t \to 0^+} \dfrac{t + 3}{t(t-1)}\)

The numerator approaches 3 and the denominator approaches 0, so the limit is undefined. (From Theorem 2.4.8)

As \(t \to 0^+\), \(t(t-1) \to 0^-\).

By part 2 of Theorem 2.6.2, as \(t \to 0^{\neq}\), \(\dfrac{1}{t(t-1)} \to -\infty\)

So, \(\lim_{t \to 0^+} \dfrac{t^2 + 3t}{t^3 - t^2} = -\infty\)

\(\lim_{x \to 1^-} \dfrac{x^2 - x + 2}{x^2 + x - 2}\)

The numerator approaches 2 and the denominator approaches 0, so the limit is undefined. (From Theorem 2.4.8)

As \(x \to 1^-\), \(x^2 + x - 2 \to 0^-\). From theore 2.6.2, as \(x \to 1^-\), \(\dfrac{1}{x^2 + x - 2} \to -\infty\).

So, \(\lim_{x \to 1^-} \dfrac{x^2 - x + 2}{x^2 + x - 2} = -\infty\)

\(\lim_{v \to 1}\dfrac{v^2 - 3v + 1}{v^2 - 2v + 1}\)

The numerator approaches -1 and the denominator approaches 0, so the limit is undefined. (From Theorem 2.4.8)

As \(v \to 1\), \(v^2 - 2v + 1 \to 0^+\). From theorem 2.6.2 as \(v \to 1\), \(\dfrac{1}{v^2 - 2v + 1} \to \infty\).

As \(v \to 1^+, v^2 - 3v + 1 \to -1^-\) As \(v \to 1^-, v^2 - 3v + 1 \to -1^+\)

\(\lim_{v \to 1}\dfrac{v^2 - 3v + 1}{v^2 - 2v + 1} = -\infty\)

\(\lim_{v \to 1}\dfrac{v^2 - 3v + 2}{v^2 - 2v + 1}\)

The numerator approaches 0 and the denominator also approaches 0.

As \(v \to 1\), \(v^2 - 2v + 1 \to 0^+\).

From theorem \(2.6.2\), as \(v \to 1\), \(\dfrac{1}{v^2 - 2v + 1} \to \infty\).

As \(v \to 1^-\), \(v^2 - 3v + 2 \to 0^+\) As \(v \to 1^+\), \(v^2 - 3v + 2 \to 0^-\)

So, the two sides limits does not exists.

\(\lim_{v \to 1^-}\dfrac{v^2 - 3v + 2}{v^2 - 2v + 1} = \infty\)

\(\lim_{v \to 1^+}\dfrac{v^2 - 3v + 2}{v^2 - 2v + 1} = -\infty\)

\(\lim_{x \to -1^+} \dfrac{x^2 + 2x - 3}{x^2 + 2|x| - 3}\)

The numerator approaches -4 and the denominator approaches 0, so the limit is undefined. (From Theorem 2.4.8)

As \(x \to -1^+\), \(x^2 + 2|x| - 3 \to 0^-\). From theorem 2.6.2 as \(x \to 1^+\), \(\dfrac{1}{x^2 + 2|x| - 3} \to -\infty\)

But the numerator is also negative. So,

\(\lim_{x \to -1^+} \dfrac{x^2 + 2x - 3}{x^2 + 2|x| - 3} = \infty\)

\(\lim_{x \to 2^-} \lfloor 4 + 3x \rfloor\)

Rewriting the above equation as \(\lim_{x \to 2^-} f(g(x))\)

where \(f(x) = \lfloor x \rfloor\), \(g(x) = 4 + 3x\)

Let \(u = g(x) = 4 + 3x\) and \(y = f(y) = \lfloor u \rfloor = \lfloor 4 + 3x \rfloor\)

\(\lim_{x \to 2^-} g(x) = 10\)

So, \(x \to 2^-\), \(u \to 10^-\)

\(\lim_{u \to 10^-} f(u) = \lim_{u \to 10^-} \lfloor u \rfloor = 9\)

\(\lim_{x \to 2^-} 4 + \lfloor 3x \rfloor\)

Rewriting the above equation as \(\lim_{x \to 2^-} f(g(x))\)

where \(f(x) = 4 + x\), \(g(x) = \lfloor 3x \rfloor\)

Let \(u = g(x) = \lfloor 3x \rfloor\) and \(y = f(y) = 4 + u\)

\(\lim_{x \to 2^-} g(x) = 5\)

\(\lim_{u \to 5} f(u) = 9\)

\(\lim_{x \to 2^-} 4 + 3\lfloor x \rfloor\)

Rewriting the above equation as \(\lim_{x \to 2^-} f(g(x))\)

where \(f(x) = 4 + 3x\), \(g(x) = \lfloor x \rfloor\)

Let \(u = g(x) = \lfloor x \rfloor\) and \(y = f(y) = 4 + 3 \lfloor u \rfloor\)

\(\lim_{x \to 2^-} g(x) = 1\)

\(\lim_{u \to 1} f(u) = 7\)

\(\lim_{x \to 0} \lfloor 1 - x^2 \rfloor\)

Rewriting the above equation as \(\lim_{x \to 0} f(g(x))\)

where \(g(x) = 1 - x^2\), \(f(x) = \lfloor x \rfloor\)

Let \(u = g(x) = 1-x^2\) and \(y = f(y) = \lfloor u \rfloor\)

\(\lim_{x \to 0} g(x) = 1-x^2 = 1\)

As \(x \to 0\), \(u \to 1^-\).

\(\lim_{u \to 1^-} f(u) = 0\)

$lim_{x → 0} ⌊ 1 - x² + x³ ⌋ $

Rewriting the above equation as \(\lim_{x \to 0} f(g(x))\)

where \(g(x) = 1 - x^2 + x^3\), \(f(x) = \lfloor x \rfloor\)

Let \(u = g(x)\) and \(y = f(y)\)

\(\lim_{x \to 0} g(x) = 1-x^2 + x^3 = 1\)

As \(x \to 0\), \(u \to 1^-\)

\(\lim_{u \to 1^-} f(u) = 0\)

$lim_{x → 0} ⌊ 1 - x² + x ⌋ $

Rewriting the above equation as \(\lim_{x \to 0} f(g(x))\)

where \(g(x) = 1 - x^2 + x\), \(f(x) = \lfloor x \rfloor\)

Let \(u = g(x)\) and \(y = f(y)\)

\(\lim_{x \to 0} g(x) = 1\)

As \(x \to 0^-\), \(u \to 1^-\)

As \(x \to 0^+\), \(u \to 1^+\)

\(\lim_{u \to 1^-} f(u) = 0\)

\(\lim_{u \to 1^+} f(u) = 1\)

Hence the limit does not exist.

\(\lim_{x \to 0} \lfloor 1 - x^2 \rfloor + x\)

\(= \lim_{x \to 0} \lfloor 1 - x^2 \rfloor + \lim_{x \to 0} x\)

\(= \lim_{x \to 0} \lfloor 1 - x^2 \rfloor\)

\(= 0\) (From solution 12)

\(\lim_{x \to 0} \dfrac{1}{x} - \dfrac{1}{x^2}\)

\(= \lim_{x \to 0} \dfrac{x-1}{x^2}\)

From theorem \(2.6.2\), the limit is \(- \infty\)

\(\lim_{v \to \infty} \dfrac{v^3 + 5}{v^2 + v}\)

Both the numerator and the denominator tend to \(\infty\). So the limit is \(\infty\).

\(\lim_{v \to -\infty} \dfrac{v^3 + 5}{v^2 + v}\)

\(= \lim_{v \to -\infty} \dfrac{v^3(1 + \dfrac{5}{v^3})}{v^2(1 + \dfrac{1}{v})}\)

\(= \lim_{v \to -\infty} \dfrac{v(1 + \dfrac{5}{v^3})}{(1 + \dfrac{1}{v})}\)

\(= -\infty\)

Suppose \(\lim_{x \to a} f(x) = L\)

Suppose \(\lim_{x \to a} g(x) = \infty\) or \(\lim_{x \to a} g(x) = -\infty\)

Case 1:

Suppose \(\lim_{x \to a} g(x) = \infty\)

From theorem 2.6.2, we know that

If \(x \to a^{\neq}\), \(g(x) \to \infty\) then as \(x \to a^{\neq}\), \(\dfrac{1}{g(x)} \to 0^+\)

So, \(\lim_{x \to a} \dfrac{1}{g(x)} = 0\)

So, \(\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a}f(x). \lim_{x \to a}\dfrac{1}{g(x)} = 0\)

Case 2:

Suppose \(\lim_{x \to a} g(x) = -\infty\)

From theorem 2.6.2, we know that

If \(x \to a^-\), \(g(x) \to -\infty\) then as \(x \to a^{\neq}\), \(\dfrac{1}{g(x)} \to 0^-\)

So, \(\lim_{x \to a} \dfrac{1}{g(x)} = 0\)

So, \(\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a}f(x). \lim_{x \to a}\dfrac{1}{g(x)} = 0\)

We need to find \(c, f\) and \(g\) such that

\(\lim_{x \to 0} f(x) = \infty\)

\(\lim_{x \to 0} g(x) = \infty\)

\(\lim_{x \to 0} \dfrac{f(x)}{g(x)} = c\)

\(f(x) = \dfrac{c}{x}\)

\(g(x) = \dfrac{1}{x}\)

\(\lim_{x \to 0} f(x) = \infty\)

\(\lim_{x \to 0} g(x) = \infty\)

\(\lim_{x \to 0} \dfrac{f(x)}{g(x)} = \dfrac{c * x}{x * 1} = c\)