We need to prove here that if \(\lim_{x \to a} f(x) = L\) then \(\lim_{x \to a^{-} f(x)} = L\) and \(\lim_{x \to a^{+}} f(x) = L\).

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

\(\forall \epsilon > 0 \; \exists \delta_1 > 0 \; \forall x a - \delta_1 < x < a + \delta_1 \implies |f(x)-L|<\epsilon\)

Let's prove \(\lim_{x \to a^{-}} f(x) = L\)

Suppose \(\delta = \delta_1 > 0\) Suppose \(a - \delta_1 < x < a\)

Since \(\delta_1 > 0\) and \(\delta_1 + a > a\). So, \(a - \delta_1 < x < a + \delta_1\). So, \(f(x) - L < \epsilon\)

Let's prove \(\lim_{x \to a^{+}} f(x) = L\)

Suppose \(\delta = \delta_1 > 0\)

Suppose \(a < x < a + \delta_1\)

We know that \(\delta_1 > 0\), So, \(-\delta_1 < 0\). So, \(a - \delta_1 < a\). So, \(a - \delta_1 < a < x < a + \delta_1\). So, we can infer that \(|f(x)-L|<\epsilon\)

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

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

\(\lim_{t \to -\infty} \dfrac{100t^2 + 1}{2t^{100} - 1}\)

\(= \lim_{t \to -\infty} \dfrac{t^2(100 + \dfrac{1}{t^2})}{t^100(2 - \dfrac{1}{t^100})} = 0\)

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

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

\(= \dfrac{0 - 1}{2 + 0} = -\dfrac{1}{2}\)

\(\lim_{x \to \infty} x^2 - \dfrac{x^4}{x^2 + 3}\)

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

\(= \dfrac{3x^2}{x^2(1 + \dfrac{3}{x^2})} = \dfrac{3}{1}\)

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

\(= \lim_{x \to \infty} \dfrac{x^2(3 - \dfrac{2}{x} - \dfrac{6}{x^2})}{x^2(1 + \dfrac{3}{x})}\)

\(= \dfrac{3}{1} = 3\)

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

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

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

\(= \dfrac{-1}{2 -0 + 0} = -\dfrac{1}{2}\)

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

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

\(= \dfrac{1 + 0}{1 - 0} = 1\)

\(\lim_{x \to -\infty} x^5 + 100x^4\)

\(= -\infty + 100 \infty\)

So, the above limits are undefined.

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

\(\lim_{x \to \infty} = \dfrac{1}{x^2 - 7x} = \lim_{x \to \infty} \dfrac{1}{x^2(1 - \dfrac{7}{x})} = 0\)

\(\lim_{x \to \infty} cos (x - 7)\) will be from \([-1,1]\).

From theorem 2.4.8, we can infer that the whole limit tends to zero.

\(\lim_{z \to -\infty} z^2 + sin z\)

\(\lim_{z \to -\infty} z^2 = \infty\)

We know that the range of \(sin x\) is \([-1,1]\). So, the limit will be \(\infty\).

\(\lim_{z \to -\infty} z(sin z + 3)\)

\(= -\infty(sin -\infty + 3) = -\infty\)

\(\lim_{x \to \infty} \dfrac{(2+x)sin^2 x}{x^2(2 + sin x)}\)

\(= \lim_{x \to \infty} \dfrac{(\dfrac{2}{2} + 1)sin^2 x}{x(2 + sin x)}\)

\(= 0\)

\(\lim_{x \to 0^+}\dfrac{x^2 - 5}{x}\)

\(= \lim_{x \to 0^+}x(1 - \dfrac{5}{x}) = -\infty\)

\(\lim_{x \to 0^-} \dfrac{1}{x^2 - 5x}\)

\(= \lim_{x \to 0^-} \dfrac{1}{x^2(1 - \dfrac{5}{x})}\)

\(= \infty\)

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

\(|u-3| = -(u-3)\)

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

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

\(\lim_{x \to 2^+} \dfrac{x + |2-x| - 2}{x^2 + 3x - 10}\)

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

\(\lim_{x \to 2^+} \dfrac{2}{x+5} = \dfrac{2}{2 + 5} = \dfrac{2}{7}\)

\(\lim_{x \to \infty} \dfrac{2x}{1 + |x|}\)

\(= \lim_{x \to \infty} \dfrac{2x}{1 + x}\)

\(= \lim_{x \to \infty} \dfrac{2}{\dfrac{1}{x} + 1}\)

\(= \dfrac{2}{0 + 1} = 2\)

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

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

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

\(= \dfrac{2}{-1} = -2\)

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

\(= \lfloor -2 + 1 \rfloor\)

\(= \lfloor -1 \rfloor\)

\(= -2\)

\(\lim_{w \to 2^-} \dfrac{w^2-4}{w - 2\lfloor w \rfloor}\)

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

\(= \lim_{w \to 2^-} \dfrac{w^2-4}{w - 2}\)

\(= \lim_{w \to 2^-} w+2\)

\(= 4\)

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

In our case,

\(f(x) = x + 2\) \(f(3 - x) = 3 - x + 2 = 4 - x\)

\(\dfrac{4 - x - 2}{x - 2} = \dfrac{2-x}{x-2} = - 1\)

Todo

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

\(\lim_{x \to 1^+} |x-1| = x - 1\)

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

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

\(\lim_{x \to -2^{+}} \dfrac{|x^2 + 2x|}{3x^2 + 5x - 2}\)

\(|x^2 + 2x| = |x(x+2)| = -x(x+2)\)

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

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

\(= \dfrac{-(-2)}{3(-2) - 1}\)

\(= \dfrac{2}{-6 - 1}\)

\(= \dfrac{-2}{7}\)

We need to prove that \(\lim_{x \to a} f(x) = \infty\)

In other words, \(\forall M \; \exists \delta > 0 \; \forall x 0 < |x-a| < \delta \implies f(x) > M\)

Suppose M.

Since \(\lim_{x \to a} g(x) = \infty\), there is some \(\delta_1 > 0\) such that \(0 < |x-a| < \delta_1 \implies g(x) > M\)

Let \(\delta = min(\delta_1, d)\). So, \(\delta > 0, \delta \leq \delta_1\) and \(\delta \leq d\).

Suppose \(0 < |x-a| < delta\). Since \(\delta \leq \delta_1\), we infer that \(0 < |x-a| < \delta_1\).

From our initial proposition, we can infer that \(g(x) > M\).

Since \(\delta \leq d\), we infer that \(0 < |x-a| < d\). So, \(f(x) \geq g(x)\).

So, \(f(x) \geq g(x) > M\). So, \(f(x) > M\).

\(\lim_{x \to -\infty} x = -\infty\)

In other words, we need to prove that

\(\forall M \; \exists N \; \forall x x < N \implies x < M\)

Suppose M be an arbitray number.

Let \(N = M\)

Suppose \(x < N\).

Since \(N = M\), we infer \(x < M\).

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

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

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

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

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

\(g(x) = |x|\)

The above functions satisfy the criteria.