Cool Proofs That You Need to Know or You'll Fail the Exam

If \(a_n\) and \(b_n\) are infinite sequences, and \(\lim_{n\to \infty}{a_n}=A\) and \(\lim_{n\to \infty}{b_n}=B\), then \(\lim_{n\to \infty}{(a_n+b_n)}=A+B\)

By laws of limits, \(\forall\varepsilon>0,\;\exists N_a,N_b\) such that \(\forall n> \text{max}(N_a,N_b)\), we have:
\(|a_n-A|<\varepsilon/2\) and \(|b_n-B|<\varepsilon/2\).
By the Triangle inequality we have:
\(|a_n+b_n-(A+B)|\le|a_n-A|+|b_n-B|<\varepsilon\)
Therefore, \(\lim_{n\to \infty}{(a_n+b_n)}=A+B\).

If \(a_n\) is a vanishing sequence, and \(b_n\) is a bounded sequence, then \(\lim_{n\to \infty}{(a_nb_n)}=0\)

Let \(\varepsilon>0\). We seek \(N\) such that if \(n>N\), \(|a_nb_n-0|<\varepsilon\).
Because \(b_n\) is bounded, there exists \(M>0\) such that \(|b_n|<M\mid\forall n\in\N\)
Thus, we have \(0\le|a_nb_n|\le M|a_n|\).
\(\lim_{n\to\infty}{a_n}=0\), so \(\exists N\) such that for \(n>N, |a_n-0|<\frac\varepsilon M\)
\(|a_nb_n|<|a_n|\cdot|b_n|<\frac\varepsilon M<\varepsilon\)
So, \(|a_nb_n|<\varepsilon\).

The derivative of sums is the sum of the derivatives: \({(f+g)}^\prime (x)=f^\prime (x)+g^\prime (x)\)

It’s the product rule! \({(f\cdot g)}^\prime (x)=(fg^\prime )(x)+(f^\prime g)(x)\)

If \(f\) is a function that is differentiable at \(x_0\), and \(x_0\) is an extreme point on \(f\), then \(f^\prime (x_0)=0\).

W.L.O.G. maximum/minimum, assuming maximum.
In order for \(f\) to be differentiable at \(x_0\), \(f^\prime _-(x_0)=f^\prime _+(x_0)\), and \(f^\prime (x_0)\) must exist.
\(f^\prime _-(x_0)=\lim_{h\to0^-}{\frac{f(x_0+h)-f(x_0)}{h}}\). Since \(f(x_0+h)-f(x_0)<0\), and \(h<0\),
\(f^\prime _-(x_0)\) is the limit of positive numbers, which we know is \(\ge0\).
\(f^\prime _+(x_0)=\lim_{h\to0^+}{\frac{f(x_0+h)-f(x_0)}{h}}\). Since \(f(x_0+h)-f(x_0)<0\), and \(h>0\),
\(f^\prime _-(x_0)\) is the limit of negative numbers, which we know is \(\le0\).
Since \(f^\prime _-(x_0)=f^\prime _+(x_0)\), \(f^\prime (x_0)\) must be zero.
\(\(Q.E.D.\)\)

If \(f\) is a function that is continuous in \([a,b]\), and differentiable in \((a,b)\), and \(f(a)=f(b)\), then \(\exists c\in(a,b)\) such that \(f^\prime(c)=0\).

There are two possibilities:

1. \(f\) is constant in \((a,b)\). Therefore, \(f^\prime (x)=0\mid\forall x\in(a,b)\).
2. \(f\) is not constant in \((a,b)\). By the Weirstrass Theorem, we know that \(\exists c\in(a,b)\) such that \(c\) is an extreme point on \(f\). By Fermat’s Theorem, \(f^\prime (c)=0\).

If \(f\) is a function that is continuous in \([a,b]\), and differentiable in \((a,b)\), then \(\exists c\in(a,b)\) such that \(f^\prime(c)=\frac{f(b)-f(a)}{b-a}\).

Let \(g(x)=f(x)-\frac{(f(b)-f(a))\cdot(x-a)}{b-a}\). Note that \(g(a)=g(b)=f(a)\). Furthermore, \(g(x)\) has the same continuity and differentiability properties as \(f(x)\), since \(g^\prime(x)=f^\prime(x)-\frac{f(b)-f(a)}{b-a}\).
Therefore, we can apply Rolle’s Theorem that \(\exists c\in(a,b)\mid g^\prime(c)=0\).
Thus, we can immediately conclude by the definition of \(g^\prime\) that \(f^\prime(c)=\frac{f(b)-f(a)}{b-a}\).

If the conditions for the MVT are met, and \(f^\prime(x)=0\mid\forall x\in(a,b)\), then \(f\) is constant in \((a,b)\).

Choose \(x_1,x_2\in(a,b)\) so that \(x_1<x_2\). By the MVT, \(\exists c\in(x_1,x_2)\) such that \(f^\prime(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}\).
By the claim, \(f^\prime(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}=0\). This can only happen if \(f(x_2)-f(x_1)=0\mid\forall x_1,x_2\in(a,b)\).

If the conditions for the MVT are met, and \(f^\prime(x)>0\mid\forall x\in(a,b)\), then \(f\) is increasing in \((a,b)\).

This is easily modified for \(f\) decreasing in \((a,b)\), and therefore omitted.

Suppose that \(f\) is not increasing in \((a,b)\). Choose \(x_1,x_2\in(a,b)\) so that \(x_1<x_2\). By the MVT, \(\exists c\in(x_1,x_2)\) such that \(f^\prime(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}\).
Since \(f\) is not increasing, and \(x_1<x_2\), \(f(x_2)-f(x_1)\le0\), and \(x_2-x_1>0\). Therefore, \(f^\prime(c)\le0\), which is a contradiction of the claim. By contradiction, \(f\) must be increasing in \((a,b)\).

If the conditions for the MVT are met for functions \(f\) and \(g\), and \(f^\prime(x)=g^\prime(x)\mid\forall x\in(a,b)\), then \(f\) and \(g\) differ by a constant—that is, \(f(x)=g(x)+C\mid\forall x\in(a,b),C\in\R\).

Define \(h(x)=f(x)-g(x)\). Then, \(h^\prime(x)=f^\prime(x)-g^\prime(x)=0\). By Corollary I, we know that if the derivative of a function is zero, that function is a constant. Therefore, \(h(x)=f(x)-g(x)=C\), which is the same thing as saying \(f(x)=g(x)+C\mid\forall x\in(a,b),C\in\R\)

If \(f\) is continuous on \([a,b]\), and \(F(x)=\int_a^xf(t)dt\) for \(a\le x\le b\), then \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(f^\prime (x)=f(x)\).

Since \(x\le c\le x+h\), and \(h\to0\), \(c=x\).
Therefore, \(f^\prime (x)=f(x)\).