Suppose \(\epsilon > 0\). Let \(\delta = \epsilon\). Suppose \(0 < |x - 0| < \delta\).

\(||x| - 0| = |x| < \delta < \epsilon\)

Suppose \(\lim_{x \to a} f(x) = 0\)

We know for \(\forall \epsilon_{1} > 0 \; \exists \delta_{1} > 0 \; \forall x \; 0 < |x-a| < \delta_{1} \implies ||f(x)| - L| < \epsilon_{1}\)

Suppose \(\epsilon > 0\).

Let \(\epsilon_{1} = \epsilon\). We know that there there \(\exists \delta_{1} > 0 \; \forall x \; 0 < |x-a| < \delta_{1} \implies ||f(x)| - L| < \epsilon_{1}\)

Let's assume \(\delta = \delta_{1}\). Suppose \(0 < |x-a| < \delta\). Since \(\delta = \delta_{1}\), we can infer that \(|f(x)| - L < \epsilon_{1}\).

Since \(\epsilon_{1} = \epsilon\), we know that \(||f(x)| - L| < \epsilon\).

In our case, we know that \(L\) is zero. So, \(||f(x)| - 0| = ||f(x)|| = |f(x)| < \epsilon\)

\(\lim{x \to 3} x^3 - 5x^2 + 7x + 2\)

\(= 3^3 - 5.3^2 + 7.3 + 2\)

\(= 5\)

\(\lim{v \to -1}v^4 + 2v^3 + 3v^2 -v + 4\)

\(= (-1)^4 + 2(-1)^3 + 3(-1)^2 -(-1) + 4\)

\(= 1 - 2 + 3 + 1 + 4\)

\(= 7\)

\(\lim_{v \to -1} v^100 + v^99 + v^98 + v^97 + ... + v^2 + v + 1\)

\(= 50 - 50 + 1\)

\(= 1\)

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

\(= \dfrac{4^2 - 12 - 4}{4 - 2}\)

\(= 0\)

\(\lim_{x \to 5}\dfrac{x^2 - 3x - 10}{x^2 - 5x}\)

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

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

\(f_1(x) = \dfrac{x^2 - 3x - 10}{x^2 - 5x}\)

\(f_2(x) = \dfrac{x + 2}{x}\)

For all \(x\) except 5, \(f_1(x) = f_2(x)\)

Also, \(0 < |x-5| < 1\), So, \(4 < x < 6\) and \(x \neq 5\). So, \(4 < x < 6\). So, \(x \neq 5\) and \(x \neq 0\). Therefore \(f_1(x)\) and \(f_2(x)\) are defined and equal.

Using theorem 2.4.5,

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

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

The above limit can be reduced to \(\lim{w \to -3} (w + 1)w\)

\(f_1(x) = \dfrac{w^3 + 4w^2 + 3w}{w + 3}\)

\(f_2(x) = (w + 1)w\)

For all \(x\) except \(-3\), \(f_1(x) = f_2(x)\)

Suppose \(0 < |w - (-3)| < 1\). We know that \(w \neq -3\). So, \(f_1(x)\) and \(f_2(x)\) are both defined. Using theorem 2.4.5,

\(\lim_{w \to -3} (w+1)w = (-3 + 1)(-3) = (-2)(-3) = 6\)

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

\(f_1(x) = \dfrac{x^2 + x - 6}{x^2 - x -2}\)

\(f_2(x) = \dfrac{x+3}{x+1}\)

Both \(f_1(x)\) and \(f_2(x)\) are same for all \(x\) except \(x = 2\) and \(x = -1\). Suppose \(0 < |x-2| < 1\). From that we know that \(x \neq 2\). Also, \(1 < x < 3\). So, \(x \neq 2\) and \(x \neq -1\). So, both \(f_1(x)\) and \(f_2(x)\) are defined and equal.

Using theorem 2.4.5,

\(\lim_{x \to 2} f_2(x) = \dfrac{5}{3}\)

\(\lim_{t \to 3}\dfrac{t^3 - 27}{t^2 - 9}\)

The above limits can be reduced to,

\(\lim_{t \to 3}\dfrac{t^2 + 9 + 3t}{t + 3}\)

From theorem 2.4.5,

\(\lim_{t \to 3}\dfrac{t^2 + 9 + 3t}{t + 3} = \dfrac{3^2 + 9 + 3^2}{6} = \dfrac{9}{2}\)

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

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

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

\(\lim_{x \to 2} x = 2\)

From theorem 2.4.8, we can conclude that the limit is undefined.

\(\lim_{x \to -1} \dfrac{ x + 4 - \dfrac{3}{x+2}}{x+1}\)

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

From theorem 2.4.5,

\(\lim_{x \to -1}\dfrac{x + 5}{x + 2} = \dfrac{5-1}{2-1} = 4\)

\(\lim_{u \to 1}\dfrac{u - \dfrac{1}{u}}{1 - \dfrac{1}{u}}\)

\(= \lim_{u \to 1}u + 1 = 2\)

\(\lim_{x \to -2}\dfrac{x}{x+2} - \dfrac{8}{x^2 - 4}\)

\(= \lim{x \to -2}\dfrac{x^3 - 16x - 16}{(x+2)(x^2 - 4)}\)

\(\lim_{x \to -2} x^3 - 16x - 16 = (-2)^3 + 32 - 16 = 8\)

\(\lim_{x \to -2} (x+2)(x^2 - 4) = 0\)

From theorem 2.4.8, we can conclude that the limit is undefined.

\(\lim_{x \to 7} \dfrac{x-8}{x-7} + \dfrac{7}{x^2 - 7x}\)

\(= \lim_{x-1}{x}\)

\(= \dfrac{6}{7}\)

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

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

\(= \lim_{z \to 0} z^2 = 0\)

\(= \lim_{z \to 0} z - 1 = -1\)

From theorem 2.4.8, we can conclude that the limit is undefined.

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

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

\(\lim_{u \to -3}[\dfrac{1}{u+3}(\dfrac{3}{u} + 1)]\)

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

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

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

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

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

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

\(\lim_{x \to 2} \dfrac{x^2 + |x| - 6}{x-2}\)

\(|x|\) can be either \(x\) or \(-x\). As \(x\) approaches 2, \(x\) will be positive. So, \(|x| = x\).

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

\(\lim_{x \to -2} \dfrac{x^2 + |x| - 6}{x+2}\)

\(|x|\) can be either \(x\) or \(-x\). As \(x\) approaches -2, \(x\) will be negative. So, \(|x| = - x\).

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

\(\lim_{x \to 2}\dfrac{x^2 + |x|-6}{x^2 + |x-4|-6}\)

\(x\) can be either \(x\) or \(-x\). As x approaches 2, \(|x|\) will b positive. So, \(|x|=x\), Similarly, \(|x-4|\) can be either \(x-4\) or \(-(x-4)\). As \(x\) approaches \(2\), \(|x-4|\) will be negative. So, \(|x-4| = -(x-4)\).

\(= \lim_{x \to 2}\dfrac{x+3}{x+1} = \dfrac{5}{3}\)

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

\(= \lfloor 2 - 0^2 \rfloor = \lfloor 2 \rfloor = 2\)

\(\lim_{x \to \dfrac{1}{2}} cos \lfloor x \rfloor\)

\(= cos \lfloor \dfrac{1}{2} \rfloor = cos \lfloor 0.5 \rfloor = cos 0 = 1\)

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

We know that \(|sin(x)| \leq 1\) for all \(x \in R\). It is also true for \(sin \dfrac{1}{x}\) for \(x \neq 0\). So,

\(|sin(\dfrac{1}{x})| <= 1\)

Since absolute values are non negative,

\(0 \leq |sin(\dfrac{x}{x})| \leq 1\)

\(0 \leq |x^2sin(\dfrac{x}{x})| \leq |x^2|\)

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

$lim_{x → 0}|x²| = 0

From squeeze theorem we can conclude that that,

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

From theorem 2.4.1, we can conclude that \(\lim_{x \to 0}sin(\dfrac{1}{x}) x^2 = 0\)

You might want to refresh the inequality properties for this question.

We know that \(0 \leq cos^2 \dfrac{\pi}{x} \leq 1\) for all \(x \neq 0\)

\(= 1 \leq cos^2 \dfrac{\pi}{x} + 1 \leq 2\)

\(= 1 \geq \dfrac{1}{cos^2 \dfrac{\pi}{x} + 1} \geq \dfrac{1}{2}\)

\(= \dfrac{1}{2} \leq \dfrac{1}{cos^2 \dfrac{\pi}{x} + 1} \leq 1\)

\(= \dfrac{|x|}{2} \leq \dfrac{|x|}{cos^2 \dfrac{\pi}{x} + 1} \leq |x|\)

Now we know that \(\lim_{x \to 0}\dfrac{|x|}{2} = 0\)

Similarly, \(\lim_{x \to 0}|x| = 0\)

Using the squeeze theorem, we can conclude that \(\lim_{x \to 0} \dfrac{|x|}{cos^2(\dfrac{\pi}{x}) + 1} = 0\)

We know that \(f(c) \neq g(c)\)

\(\forall x \neq c f(x) = g(x)\)

We need to prove \(\forall a \lim_{x \to a} f(x) = \lim_{x \to a} g(x)\)

Suppose \(a\) be an arbitrary element. Let us consider the cases:

Case 1: \(a \neq c\)

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

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

Since \(a \neq c\), \(f(a) = f(a)\) and hence \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x)\)

Case 2: \(a = c\)

\(\lim_{x \to a}f(x)\)

Nowte that in the above limit \(x \neq a\) but close to the value of a. So, \(\lim_{x \to a} f(x)\) is some value \(L\). Similarly, since \(\forall x \neq c f(x) = g(x)\) we can conclude that \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x)\).

Suppose \(\epsilon > 0\). We need to find \(\delta > 0\) such that \(0 < |x-a| < \delta \implies |f(x) -L | < \epsilon\)

Since \(\lim_{x \to a} g(x) = L\), by definition of limits there exists some \(\delta_1 > 0\) such that

\(0 < |x - a| < \delta_1 \implies |g(x) - L | < \epsilon\)

\(0 < |x - a| < \delta_1 \implies L - \epsilon < g(x) < L + \epsilon\)

Similarly, since \(\lim_{x \to a} h(x) = L\), we know that

\(0 < |x - a| < \delta_2 \implies L - \epsilon < h(x) < L + \epsilon\)

Additionally, since \(g(x) \leq f(x) \leq h(x)\) in \(forall x \in (b,c)\) and \(x \neq a\). So,

\(0 < |x - a| < \delta_3 \implies g(x) \leq f(x) \leq h(x)\)

Suppose \(\delta = min(\delta_1, \delta_2, \delta_3)\). Then since \(\delta \leq \delta_1\), \(\delta \leq \delta_2\) and \(\delta \leq \delta_2\), we can infer that

\(L - \epsilon < g(x) < L + \epsilon\)

\(L - \epsilon < h(x) < L + \epsilon\)

\(g(x) \leq f(x) \leq h(x)\)

Combining them we get,

\(L - \epsilon < g(x) \leq f(x) \leq h(x) < L + \epsilon\)

Therefore, \(-\epsilon < f(x) - L < \epsilon\) which is \(|f(x) - L| < \epsilon\)

You can find a similar proof here.

Let \(L = \lim_{x \to a}g(x)\) Suppose \(\lim_{x \to a} f(x) + g(x)\) is defined and the limit is \(M\). Let's assume \(\epsilon = \dfrac{\epsilon}{2}\) for both the cases. So,

\(0 < |x -a| < \delta_1 \implies |f(x) + g(x) -M| < \dfrac{\epsilon}{2}\)

\(0 < |x -a| < \delta_2 \implies |g(x) - L| < \dfrac{\epsilon}{2}\)

Suppose \(\epsilon > 0\)

Let \(\delta = min(\delta_1, \delta_2)\). So, \(\delta > 0\), \(\delta \leq \delta_1\) and \(\delta \leq \delta_2\)

We can infer from the above equation,

\(|f(x) + g(x) -M| < \dfrac{\delta}{2}\)

\(|g(x) - L| < \dfrac{\delta}{2}\)

From the inequality laws we know that, if \(a < b\) and \(c < b\), then \(a + c < 2b\).

So, \(|f(x) + g(x) -M| + |g(x) - L| < \epsilon\)

\(|f(x) -M + L| = |f(x) + g(x) -M - (g(x) - L)|\)

\(<= |f(x) + g(x) -M| + |g(x)-L|\) (From triangle inequality)

\(< \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} < \epsilon\)

The above conclusion means,

\(\forall \epsilon > 0 \; \exists \delta = min(\delta_1, \delta_2) \; \forall x 0 < |x-a| < \delta \implies |f(x) + L - M| < \epsilon\)

So, the limit of \(f(x)\) is defined. But we know that \(f(x)\) is undefined which is a contradiction. Hence, \(\lim_{x \to a} f(x) + g(x)\) is undefined.