Let \((x_n)\) be increasing and bounded and \(x = \sup(x_n, n \in \mathbb{N})\)

Let \(\varepsilon > 0\),

then \(x - \varepsilon\) is not upper bound.

You can find \(N\) s.t. \(x - \varepsilon < x_n \leq x\)

Since \((x_n)\) increasing, \(x \geq x_N\)

Let \((-x_n)\) be increasing and bounded

So \[\lim_{n\to\infty} (-x_n) = \sup(-x_n) = -\inf(x_n)\]

So \[\lim_{n\to\infty} = \inf(x_n)\]

Let \[x = \lim_{x \to \infty} x_n\]

Let \(\varepsilon > 0\). We can find \(f(\varepsilon)\) s.t. \(|x_n - x| \leq \varepsilon\) for \(n \geq f(\varepsilon)\)

Since \(n_k\) is strictly increasing, \(n_k \geq k\)

For \(k \geq f(\varepsilon), n_k \geq k \geq f(\varepsilon)\)

so \(|x_{n_k} - x| \leq \varepsilon\) and \[\lim_{k\to\infty} x_{n_k} = x\]

Let \((x_n)\) be a sequence.

Then \((x_n)\) does not converge if we can find \(\varepsilon_0 > 0\) s.t. \(\forall K \geq 0, \exists n_K \geq K, |x_{n_K} - x| > \varepsilon_0\)

So if \((x_n)\) does not converge to \(x\), we can find \(\varepsilon_0 > 0\) and a subsequence \((x_m)\) s.t. \(\forall k \in \mathbb{N}, | x_m - x | > \varepsilon_0\)

Define \(x_m\) to be a peak if \(x_n \leq x_m\) for \(n \geq m\)

Two cases:

1. \((x_n)\) has infinitely many peaks

   Let \[x_{n_1},x_{n_2},x_{n_3},...\] be a sequence of peaks.

   Then \[x_{n_1} \geq x_{n_2} \geq x_{n_3} \geq ...\]

   So the subsequence is increasing

2. \((x_n)\) has finite number of peaks

   Then you can find \(N\) s.t. \(\forall n \geq N\), \(x_n\) is not a peak

   Then the subsequence \[x_{n_N},x_{n_{N+1}},x_{n_{N+2}},...\] must be increasing (they are not peaks)

Let \(a,b \in \mathbb{R}\) s.t. \(\forall n \in \mathbb{N}, x_n \in [a,b]\)

Then either \([a,\frac{a+b}{2}]\) or \([\frac{a+b}{2},b]\) is contains an infinite subsequence of \((x_n)\). Call this \(I_1\).

By induction, construct \(I_n\)

• width of \(I_n = \frac{b-a}{2}\)
• \(I_{n+1} \in I_n\)
• \(\{k \in \mathbb{N} | x_k \in I_n \}\) is infinite

We know \(\cap I_n\) is nonempty

Let \(x \in \cap I_n\)

We can construct subsequence \(x_{n_k}\) such that \(x_{n_k} \in I_k\)

Since \(x \in I_k\) and \(| x_{n_k} - x| \leq \frac{b-a}{2}\)

\[\lim_{k \to \infty} x_{n_k} = x\]

Let \(\varepsilon > 0\).

Let \[x = \lim_{n \to \infty} x_n\]

\(\exists K(\varepsilon) \geq 0\) such that \(\forall n \geq K(\varepsilon)\) \(|x_n - x| \leq \frac{\varepsilon}{2}\)

For \(n,m \geq K(\varepsilon)\),

\[ \begin{aligned} \ |x_n - x_m| & = |x_n - x + x - x_m| \\ & \leq |x_n + x| + |x - x_m| \\ & \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \\ \end{aligned} \]

So \((x_n)\) is a Cauchy sequence

Let \((x_n)\) be a Cauchy sequence. Then \((x_n)\) is bounded and has a convergent subsequence \((x_{n_k})\)

s.t. \[\lim (x_{n_k}) = x\]

Let \(\varepsilon \geq 0\).

There exists \(K(\varepsilon) \geq 0\) s.t. \(\forall n,m \geq K(\varepsilon), |x_n - x_m| \leq \varepsilon\)

there exists \(N \geq K(\varepsilon), |x - x_n| \leq \varepsilon\)

For \(n \geq K(\varepsilon)\)

\[ \begin{aligned} \ |x_n - x| & = |x_n - x_N + x_N - x| \\ & \leq |x_n - x_N| + |x_N - x| \\ & \leq \varepsilon + \varepsilon = 2 \varepsilon \end{aligned} \]

Let \((x_n)\) be a contracting sequence

\[ \begin{aligned} \ | x_{n+1} - x_n| & \leq C |x_n - x_{n-1}| \\ & \leq C^2 | x_{n-1} - x_{n-2} | \\ & \leq C^3 | x_{n-1} - x_{n-2} | \\ & \leq ... \\ & \leq C^{n-1} | x_2 - x_1 | \end{aligned} \]

\[ \begin{aligned} \ |x_{n+m} - x_n| & = |x_{n+m} - x_{n+m-1} + ... + x_{n+1} - x_n| \\ & \leq |x_{n+m} - x_{n+m-1}| + ... + |x_{n+1} - x_n| \\ & \leq C^{n+m-1} |x_{n+m} - x_{n+m-1}| + ... + C ^{n-1} |x_{n+1} - x_n| \\ & \leq C^{n-1} (1 + ... + C^m) |x_2 - x_1| \\ & \leq C^{n-1} \frac{1-C^{n+1}}{1-C} |x_2 - x_1| \\ & \leq \frac{C^{n-1}}{1-C} |x_2 - x_1| \\ \end{aligned} \]

\[\lim_{n \to \infty} C^{n-1} = 0\]

So \((x_n)\) is a Cauchy sequence and is convergent

Let \((x_n)\) be increasing and not bounded Let \(M \geq 0\), Since \((x_n)\) is not bounded, you can find \(n \geq 0\) s.t. \(x_n \geq M\) For \(k \geq n, x_k \geq x_n\) so \(x_k \geq x_n \geq M\)

If \((x_n)\) is decreasing and bounded, \((-x_n)\) is increasing and not bounded,

so \[\lim_{n \to \infty} -x_n = \infty\]

\[\lim_{n \to \infty} x_n = -\infty\]

Let \(M \geq 0\)

We can find \(N \geq 0\) s.t. for \(n \geq N\), \(x_n \geq M\)

So \(y_n \geq M\)

\[\lim_{n \to \infty} y_n = \infty\]

Since \(x_n = s_n - s_{n-1}\)

\[\lim_{n \to \infty} x_n = \lim_{n \to \infty} s_n - \lim_{n \to \infty} s_{n-1} = 0\]

So \((x_n)\) is convergent

Since \((x_n)\) is positive, \((s_n)\) is increasing.

Increasing series are either bounded and convergent

or unbounded and divergent

\(0 \leq x_1 \leq y_1\)

\(0 \leq x_2 \leq y_2\)

…

\(0 \leq x_n \leq y_n\)

So, \(0 \leq s_n \leq t_n\)

Assume \((t_n)\) is convergent.

Then \((t_n)\) is bounded, then \((s_n)\) is bounded.

Since \((s_n)\) is increasing and bounded, it is convergent

Assume \((s_n)\) is divergent.

then \((s_n)\) is unbounded, then \((t_n)\) is unbounded

Since \((s_n)\) is increasing and unbounded, it is divergent

Let \(\varepsilon > 0\)

Since \(L\) is a limit, \(\exists \zeta > 0\) s.t.

\(|x - a| \leq \zeta \Rightarrow |f(x) - L| \leq \varepsilon\)

Since \(L'\) is a limit, \(\exists \zeta' > 0\) s.t.

\(|x - a| \leq \zeta' \Rightarrow |f(x) - L'| \leq \varepsilon\)

Let \(x \in \mathbb{R}\) s.t.

\(|x - a| \leq \min(\zeta, \zeta')\)

Then

\[ \begin{aligned} \ |L - L'| & \leq |L - f(x) + f(x) - L'| \\ & \leq |L - f(x)| + |f(x) - L| \\ & \leq \varepsilon + \varepsilon \leq 2 \varepsilon \end{aligned} \]

We must show that 1 implies 2 and 2 implies 1.

1 -> 2

Assume that 1 is true. Let \(x_n\) be a sequence s.t. \(x_n \neq c\) and \(\lim_{n \to \infty} x_n = c\)

Let \(\varepsilon > 0\). We can find \(\zeta > 0\) s.t.

for \(x \in (c - \zeta, c + \zeta), x \neq c\) we have \(|f(x) - L| \leq \varepsilon\)

Since \(x_n\) converges to \(c\), we can find \(K\) s.t.

for \(n \geq K\), \(x_n \in (c - \zeta, c + \zeta)\)

For \(n \geq K, |f(x_n) - L| \leq \varepsilon\)

2 -> 1

Assume for contradiction that 2 is true, and 1 is not true.

There exists \(\varepsilon_0 > 0, \forall \zeta > 0\)

\(\exists x \in (c - \zeta, c + \zeta), x \neq c\)

\(|f(x) - L| > \varepsilon_0\)

So, \(\exists x _n \in (c - \frac{1}{n}, c + \frac{1}{n}), x_n \neq c\)

\(|f(x_n) - L| > \varepsilon_0\)

We have that \(\lim_{n \to \infty} x_n = c\), but \(\lim_{n \to \infty} f(x_n) \neq L\)

This contradicts 2

Let \(L = \lim_{x \to 1} f(x)\). We can find \(\zeta > 0\)

s.t. for \(x \in [c - \zeta, c + \zeta] \cap A, x \neq c\),

\(|f(x) - L| \leq 1\)

We have \(|f(x)| \leq 1 + |L|\)

Let \((x_n)\) be a sequence in \(A\), s.t. \(\lim_{n \to \infty} x_n = c\) and \(x_n \neq c\)

Since \(\lim_{n \to \infty} x_n\) exists, the sequence \(f(x_n)\) is convergent and \(\lim_{n \to \infty} f(x_n) = \lim_{n \to c} f(x)\)

similarly, \(\lim_{n \to \infty} g(x_n) = \lim_{n \to c} g(x)\)

so, \(f(x_n) + g(x_n)\) is convergent and \(\lim_{n \to \infty} f(x_n) + g(x_n) = \lim_{n \to \infty}f(x_n) + \lim_{n \to \infty}g(x_n) = \lim_{n \to c}f(x) + \lim_{n \to c}g(x)\)

Let \((x_n) \in A\) be a sequence which converges to \(c\), but \(x_n \neq c\)

\(\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} h(x_n) = L\)

From the squeeze theorem for sequences, \(g(x_n)\) converges and \(\lim_{n \to \infty} g(x_n) = L\)

1 -> 2

Assume that \(\lim_{x \to c} f(x) = \infty\)

Let \(M > 0\). We can find \(\zeta > 0\)

s.t. for \(x \in [c - \zeta, c + \zeta] \cap A, f(x) > M, x \neq c\)

Since \(x_n\) converges to \(c\), we can find \(N \geq 0\), s.t. for \(n \geq N, x_n \in [c - \zeta, c + \zeta]\) and \(f(x_n) \geq M\)

so \(\lim_{n \to \infty} f(x_n) = \infty\)

2 -> 1

Assume for contradiction that \(\lim_{x \to c} f(x) \neq \infty\). Then we can find \(M > 0\) s.t. \(\forall \zeta\),

there exists \(x \in [c - \zeta, c + \zeta] \cap A, x \neq c\) where \(f(x) \leq M\)

Let \(\zeta = \frac{1}{n}\). \(x_n \in [c - \zeta, c + \zeta], f(x_n) \leq M\)

we don't have \(\lim_{n \to \infty} f(x_n) = \infty\) which is a contradiction

Let \((x_n) \in A\) converge to \(c\).

Since \(f,g\) are continuous at \(c\),

\(\lim_{n \to \infty} f(x_n) = f(c)\) and \(\lim_{n \to \infty} g(x_n) = g(c)\)

so \(\lim_{n \to \infty} f(x_n) + \lim_{n \to \infty} g(x_n) = f(c) + g(c)\)

The proof is identical for \(fg\) and \(\frac{f}{g}\)

Let \(f: I \to \mathbb{R}\). \(S = \sup f\)

We can find a sequence \(x_n \in I\) s.t. \(f(x_n)\) converges to \(S\).

We can find a convergfent subsequence \(x_m\).

Let \(x^* = \lim x_m\)

Since \(f\) is continuous at \(x^*\),

\(\lim f(x_m) = f(x^*)\)

Therefore \(S = f(x^*)\)

Let \(A = {x \in [a,b], f(x) \leq \lambda}\)

\(A \neq crosszero\) and is bounded so \(A\) has an upper bound

Let \(c = \sup A\)

Let \((x_n)\) be a sequence in \(A\), s.t. \(\lim x_n = c\)

Since \(x_n \in A\), f(x_n) ≤ λ$

Since \(f\) is continuous at \(c\), \(\lim f(x_n) = f(c)\)

Therefore \(f(c) \leq \lambda\)

For \(n \geq 1\), \(c + \frac{1}{n} \notin A\), so \(f(x + \frac{1}{n} > \lambda\)

So \(f(c) \geq \lambda\)

We conclude \(f(c) = \lambda\)

Let \(I\) be an interval. Let \(a,b\) s.t. \(a \leq b\) and \(a,b \in f(I)\)

Since \(a \in f(I), \exists \alpha \in I\) s.t.

\(f(\alpha) = a\)

Since \(b \in f(I), \exists \beta \in I\) s.t.

\(f(\beta) = b\)

For \(c \in [a,b]\) we can find \(\gamma \in [\alpha, \beta]\) s.t. \(f(\gamma) = c\)

So \(f(\gamma) = c\). Therefore \(f(I)\) is an interval

Let \(\varepsilon > 0\). For \(x,y \in A\) such that \(|x - y| \leq \frac{\varepsilon}{K}\)

we have \(|f(x) - f(y)| \leq \varepsilon\)

Let \(f:I \to \mathbb{R}\) be a differentiable function.

Assume \(|f'(x)| \leq K\) for all \(x \in I\)

Using the Mean Value Theorem:

There exists a \(c \in I\) such that \(\frac{f(x) - f(y)}{x - y} = f'(c)\)

if we let \(K = f'(c)\), then \(|f(x) - f(y)| \leq K|x - y|\)

Let \(f\) be increasing. \(\varepsilon > 0\), and \(c \in \mathbb{R}\)

Let \(L = \sup\{f(x)| x < c\}\)

\(L - \varepsilon\) is not an upper bound, so

there exists \(x_{\varepsilon} < c\) such that \(f(x_{\varepsilon}) > L - \varepsilon\).

For \(x \in [x_{\varepsilon}, c)\)

\(L - \varepsilon \leq f(x) \leq L\)

so \(\lim_{x \to +c} f(x) = L\), \(\lim_{x \to -c} f(x) = L\).

The proof is similar for right limits and infimums

Let \(x,y \in A\) such that \(x \neq y\)

Since \(f\) is increasing, we have \(x < y\) or \(x > y\), so

\(f(x) < f(y)\) or \(f(x) > f(y)\), so \(f(x) \neq f(y)\)

so \(f\) is injective.

Let \(y_1,y_2 \in f(A)\) such that \(y_1 < y_2\)

We have \(y_1 = f(x_1), y_2 = f(x_2)\) for some \(x_1,x_2 \in A\)

We must have \(x_1 < x_2\), so \(f^{-1}{y_1} < f^{-1}{y_2}\)

Let \(f(I) = J\)

Assume for contradiction that \(g\) is not continuous.

Then \(g(J) = g(f(I)) = I\) is not an interval, which is a contradiction

Let \(f:I \to \mathbb{R}\) be differentiable. Then \(\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = L\)

\(\frac{f(x) - f(c)}{x - c}(x - c) = L(x-c)\)

so \(\lim_{x \to c} f(x) - f(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \times \lim_{x \to c} (x-c) = f'(c) \times 0 = 0\)

so \(\lim_{x \to c} f(x) = f(c)\)

\[(f + g)'(c) = \frac{f(x) + g(x) - (f(c) + g(c)) }{x - c} = \frac{f(x) - f(c)}{x - c} + \frac{g(x) - g(c)}{x - c}\] so

\[\lim_{x \to c} (f + g)'(c) = f'(c) + g'(c)\]

\[\frac{f(x)g(x) - f(c)g(c)}{x - c} = \frac{f(x)g(x) - f(x)g(c) + f(x)g(c) - f(c)g(c)}{x - c} = f(x) \frac{g(x) - g(c)}{x - c} + g(c) \frac{f(x) - f(c)}{x - c}\]

so \(\lim_{x \to c} \frac{f(x)g(x) - f(c)g(c)}{x - c} = f'(c)g(c) + f(c)g'(c)\)

\[\frac{\frac{1}{f(x)} - \frac{1}{f(c)}}{x - c} = \frac{f(c) = f(x)}{f(x)f(c)(x-c)}\]

so

\(\lim_{x \to c} \frac{\frac{1}{f(x)} - \frac{1}{f(c)}}{x - c} = \frac{-f'(c)}{f(c)^2}\)

\(\left( \frac{f}{g} \right)(x) = f(x) \frac{1}{g(x)}\)

so \[\left(\frac{f}{g}\right)' (c) = f'(c) \times \frac{1}{g(x)} + f(c) \times \left(\frac{1}{g}\right)'(c) = \frac{f'(c)}{g(c)} - f(c) \frac{g'(c)}{g(c)^2} = \frac{f'g(c) - fg'(c)}{g(c)^2}\]

Assume \(f\) is differentiable at \(c\). Let

\[ \phi(x) = \begin{cases} \frac{f(x) - f(c)}{x - c} & x \neq c \\ f'(c) & x = c \end{cases} \]

Then \(\phi\) is continuous and \(f(x) - f(c) = \phi(x)(x - c)\) for all \(x \in I\)

Assume \(\phi:I \to \mathbb{R}\) is continuous such that \(f(x) - f(c) = \phi(x)(x - c)\) for all \(x \in I\).

\(f(x) - f(c) = \phi(x)(x - c) \rightarrow \frac{f(x) - f(c)}{x - c} = \phi(x)\)

\(\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c}\phi(x) = \phi(c)\)

so \(f\) is differentiable at \(c\) and \(f'(c) = \phi(c)\)

By the Carathéoday Lemma,

\(f(x) - f(c) = \phi(x)(x - c), x \in I\)

\(g(y) - g(f(c)) = \Psi(y)(y - f(c)), y \in f(I)\)

So \(g(f(x)) - g(f(c)) = \Psi(f(x))(f(x) - f(c)) = \Psi(f(x))(\phi(x)(x - c))\)

and \(\lim_{x \to c} \frac{g(f(x)) - g(f(c))}{x - c} = \Psi(f(c))\phi(f(c)) = g'(f(c))f'(c)\)

Since \(f\) is continuous and differentiable at \(c\), we can find a continuous \(\phi\) such that \(f(x) - f(c) = \phi(x)(x - c)\).

Let \(d = f(c) \in f(I)\) where \(f'(c) \neq = 0\)

\(x = g(y), y \in f(I)\)

\(f(g(y)) - f(g(d)) = \phi(g(y))(g(y) - g(d))\)

\(y - d = \phi(g(y))(g(y) - g(d))\)

\(\phi(g(d)) = f'(c) \neq 0\)

Since \(\phi(g(y))\) is continuous so \(\phi(g(y)) \neq 0\) in a neighborhood around \(d\).

so \(\lim_{y \to d} \frac{g(y) - g(d)}{y - d} = \frac{1}{\phi(g(d))} = \frac{1}{f'(g(d))}\)

Assume for contradiction \(f'(x_0) > 0\)

We know \(\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = f'(x_0)\)

So there exists \(\zeta > 0\) such that

for \(x \in [x_0 - \zeta, x_0 + \zeta] \cap I\), we have \(\frac{f(x) - f(x_0)}{x - x_0} > 0 \longrightarrow f(x) > f(x_0)\)

This contradicts that \(x_0\) is a relative minimum.

The proof can be repeated for relative maximum.

Since \(f\) is differentiable, it is continuous. Since \([a,b]\) is closed and bounded we can find \(c,d \in [a,b]\),

such that \(f(c) = \sup(f(x)\) for \(x \in [a,b]\)

\(f(d) = \inf(f(x)\) for \(x \in [a,b]\)

If \(c\) and \(d\) are both boundary points, then \(f\) must be 0, so \(f'(x) = 0\) for all \(x\).

If one of \(c\) or \(d\) is not a boundary point, then the derivative at this relative extrema is 0.

So \(f\) has at least one point between \(a\) and \(b\) where the derivative is 0.

We will make use of Rolle's theorem by transforming the function so that the tangent line is flat.

Let \(g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a)\)

So \(g(a) = g(b) = 0\)

\(g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}\)

Then there exists \(c \in (a,b)\) such that \(g'(c) = 0\)

so \(f'(c) = \frac{f(b) - f(a)}{b - a}\)

Let \(a,b \in I\) such that \(a < b\)

From the mean value theorem, we can find \(c \in I\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a} = 0\)

\(f(b) - f(a) = 0 \longrightarrow f(b) = f(a)\)

So \(f\) is constant

Using the definition of derivative:

If \(f\) is increasing, then \(f'(x) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}\).

since \(\frac{f(x) - f(c)}{x - c}\) is a positive function,

therefore, \(f'(x) \geq 0\)

Assume for all x, \(f'(x) \geq 0\)

From the MVT, we can find \(c \in (a,b)\), such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\)

Since \(f'(c) \geq 0\), \(f\) is increasing.

\[\frac{f(x)}{g(x)} = \frac{\frac{f(x) - f(a)}{x - a}}{\frac{g(x) - g(a)}{x - a}}\]

so \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}\)

Let \(h(x) = \frac{f(b) - f(a)}{g(b) - g(a)} \left[g(x) - g(a)\right] - \left[f(x) - f(a)\right]\)

then \(h(a) = 0\)

\(h'(x) = \frac{f(b) - f(a)}{g(b) - g(a)} g'(x) - f'(x)\)

From Rolle's theorem, there exists \(c\) such that \(h'(c) = 0\)

\(\frac{f(b) - f(a)}{g(b) - g(a)} g'(x) - f'(x) = 0\)

so \[\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}\]

Proof for L'Hôpital's Rule #2

Let \(f,g:(a,b) \to \mathbb{R}\) be differentiable functions.

Assume \(\lim_{x \to a+} f(x) = \lim_{x \to a+} g(x) = 0\)

Assume \(g'(x) \neq\) for \(x \in (a,b)\)

a) Assume \(\lim_{x \to a+} \frac{f'(x)}{g'(x)} = L\)

Let \(\varepsilon > 0\) We can find \(\delta > 0\) such that for \(x \in (a, a + \delta)\)

we have \(L - \varepsilon \leq \frac{f'(x)}{g'(x)} \leq L + \varepsilon\)

Let \(\alpha, \beta \in (a, a + \delta)\) where \(\alpha < \beta\)

We can find \(\gamma \in (\alpha, \beta)\) such that \(\frac{f(\beta) - f(\alpha)}{g(\beta) - g(\alpha)} = \frac{f'(\gamma)}{g'(\gamma)}\)

so \(L - \varepsilon \leq \frac{f(\beta) - f(\alpha)}{g(\beta) - g(\alpha)} \leq L + \varepsilon\)

Take the limit for \(\alpha \to a\)

Then \(L - \varepsilon \leq \frac{f(\beta)}{g(\beta)} \leq L + \varepsilon\), for \(\beta \in (a, a + \delta)\)

so \(\lim_{x \to a+} \frac{f(x)}{g(x)} = L\)

b) Assume \(\lim_{x \to a+} \frac{f'(x)}{g'(x)} = \infty\)

Let \(M \geq 0\).

There exists \(\delta > 0\) such that \(\frac{f'(x)}{g'(x)} \geq M\) for \(x \in (a, a + \delta)\)

we can find \(\gamma \in (\alpha, \beta)\) such that \(\frac{f(\beta) - f(\alpha)}{g(\beta) - g(\alpha)} = \frac{f'(\gamma)}{g'(\gamma)}\)

so \(\frac{f(\beta) - f(\alpha)}{g(\beta) - g(\alpha)} \geq M\)

Let \(\alpha \to a\). Then \(\frac{f(\beta)}{g(\beta)} \geq M\)

so \(\lim_{x \to a+} \frac{f(x)}{g(x)} = \infty\)

1. Since \(S(f+g,\bar{\cal{P}}) = S(f,\bar{\cal{P}}) + S(g,\bar{\cal{P}})\) (linearity of sums and limits) we have \[\begin{aligned} \lim_{||\bar{\cal{P}}|| \to 0} S(f+g,\bar{\cal{P}}) &= \lim_{||\bar{\cal{P}}|| \to 0} S(f,\bar{\cal{P}}) + \lim_{||\bar{\cal{P}}|| \to 0} S(g,\bar{\cal{P}})\\ \int_a^b f(x) + g(x) dx &= \int_a^b f(x) dx + \int_a^b g(x) dx \end{aligned} \]

   A similar proof can be performed for \(S(kf,\bar{\cal{P}}) = kS(f+g,\bar{\cal{P}})\)

2. Proof idea: Riemann sums cannot be defined for unbounded functions