4.5 Solutions

Convergence

Theorem. Suppose $x_n \to x$ and $y_n \to y$. Then $x_n + y_n \to x + y$.

Proof. Let $\epsilon>0$. $$\begin{alignat*}{2} \text{\textcircled{1}} \quad \exists N_x: n>N_a \implies \lvert x_n - x\rvert &< \tfrac{\epsilon}{2} &\hspace{4em}& x_n \to x \\ \text{\textcircled{2}} \quad \exists N_y: n>N_y \implies \lvert y_n - y\rvert &< \tfrac{\epsilon}{2} && y_n \to y \end{alignat*}$$ Thus, for all $n>N=\max(N_x,N_y)$, we have $$\begin{alignat*}{2} \lvert x_n + y_n - (x+y)\rvert &\le \lvert x_n - x\rvert + \lvert y_n - y\rvert &\hspace{6em}& \text{Triangle inequality} \\ &< \epsilon&& \text{\textcircled{1}}, \text{\textcircled{2}} \end{alignat*}$$  

Theorem. Suppose $x_n \to x$ and $y_n \to y$. Then $x_n y_n \to xy$.

Proof. First, let’s establish an identity: $$\begin{align*} x_n y_n - xy &= x_n y_n - x_n y + x_n y - xy \\ &= x_n (y_n - y) + y(x_n - x) \\ &= (x_n - x + x)(y_n - y) + y(x_n - x) \\ &= (x_n - x)(y_n - y) + x(y_n -y) + y(x_n - x) \end{align*}$$
Now, let $\epsilon>0$. $$\begin{alignat*}{2} \text{\textcircled{1}} \quad \exists N_x: n>N_x \implies &\lvert x_n - x\rvert < \frac{\sqrt{\epsilon}}{3} + \frac{\epsilon}{3\lvert y\rvert} &\hspace{4em}& x_n \to x \\ \text{\textcircled{2}} \quad \exists N_y: n>N_y \implies &\lvert y_n - y\rvert < \frac{\sqrt{\epsilon}}{3} + \frac{\epsilon}{3\lvert x\rvert} && y_n \to y \end{alignat*}$$ Thus, for all $n>N=\max(N_x,N_y)$, we have $$\begin{alignat*}{2} \lvert x_n y_n - xy\rvert &= \lvert(x_n - x)(y_n - y) + x(y_n -y) + y(x_n - x)\rvert &\hspace{4em}& \text{Identity above} \\ &\le \lvert x_n-x\rvert\lvert y_n-y\rvert + \lvert x\rvert\lvert y_n-y\rvert + \lvert y\rvert\lvert x_n-x\rvert && \text{Triangle inequality} \\ &< \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} && \text{\textcircled{1}}, \text{\textcircled{2}} \\ &= \epsilon \end{alignat*}$$ In the construction of $N$ above, note that we are assuming $x,y \ne 0$. If either is zero, the second term in the sum can simply be omitted, as the corresponding term below is zero.

Theorem. Suppose $x_n \to x$, with $x_n \ne 0$ for all $n$ and $x \ne 0$. Then $1/x_n \to 1/x$.

Proof.

First, let us note that $\lvert a-b\rvert < \tfrac{1}{2}\lvert b\rvert \implies \lvert a\rvert > \tfrac{1}{2}{b}$. This is fairly obvious when you think about it; to prove it, we can break the claim up into cases:

The cases where $b<0$ follow the same reasoning. Now, let $\epsilon>0$. $$\begin{alignat*}{3} \text{\textcircled{1}} &\quad& \exists N_1: n>N_1 \implies \lvert x_n - x\rvert &< \tfrac{1}{2}\lvert x\rvert^2\epsilon&\hspace{8em}& x_n \to x \\ \text{\textcircled{2}} && \exists N_2: n>N_2 \implies \lvert x_n-x\rvert &< \tfrac{1}{2}\lvert x\rvert && x_n \to x \\ \text{\textcircled{3}} && \text{so that } \lvert x_n\rvert &> \tfrac{1}{2}\lvert x\rvert && \text{\textcircled{2}}, \text{see above} \end{alignat*}$$ Thus, for all $n>N=\max(N_1, N_2)$, we have $$\begin{alignat*}{2} \lvert\frac{1}{x_n} - \frac{1}{x}\rvert &= \lvert\frac{x-x_n}{x_n x}\rvert \\ &\le \frac{2}{\lvert x\rvert^2}\lvert x_n-x\rvert &\hspace{8em}& \text{\textcircled{3}} \\ &< \epsilon&& \text{\textcircled{1}} \end{alignat*}$$ Note that in this third theorem, the requirement that $x_n \ne 0$ is unnecessary. As we see from $\text{\textcircled{3}}$, if $x_n \to x$ and $x \ne 0$, then there is an $N$ such that $x_n \ne 0$ for all $n>N$.

Continuity

The first two theorems are essentially the same as their sequence counterparts, but the differences are worth paying attention to.

Theorem. Let the functions $f$ and $g$ be continuous at $x_0$. Then $h = f + g$ is continuous at $x_0$.

Proof. Let $\epsilon>0$. $$\begin{alignat*}{2} \text{\textcircled{1}} \quad \exists \delta_f: \lvert x-x_0\rvert < \delta_f &\implies \lvert f(x) - f(x_0)\rvert < \tfrac{\epsilon}{2} &\hspace{6em}& f \text{ continuous at } x_0 \\ \text{\textcircled{2}} \quad \exists \delta_g: \lvert x-x_0\rvert < \delta_g &\implies \lvert g(x) - g(x_0)\rvert < \tfrac{\epsilon}{2} && g \text{ continuous at } x_0 \end{alignat*}$$ Thus, for all $x:\lvert x-x_0\rvert < \delta=\min(\delta_f, \delta_g)$, we have $$\begin{alignat*}{2} \lvert h(x) -h(x_0)\rvert &= \lvert f(x) + g(x) - f(x_0) - g(x_0)\rvert &\hspace{6em}& \text{Def } h \\ &\le \lvert f(x)-f(x_0)\rvert + \lvert g(x)-g(x_0)\rvert && \text{Triangle inequality} \\ &\le \epsilon&& \text{\textcircled{1}}, \text{\textcircled{2}} \end{alignat*}$$  

Theorem. Let the functions $f$ and $g$ be continuous at $x_0$. Then $h = f \cdot g$ is continuous at $x_0$.

Proof. Let $\epsilon>0$. $$\begin{alignat*}{2} \text{\textcircled{1}} \quad \exists \delta_f: \lvert x-x_0\rvert < \delta_f &\implies \lvert f(x) - f(x_0)\rvert < \frac{\sqrt{\epsilon}}{3} + \frac{\epsilon}{3\lvert g(x_0)\rvert} &\hspace{3em}& f \text{ continuous at } x_0 \\ \text{\textcircled{2}} \quad \exists \delta_g: \lvert x-x_0\rvert < \delta_g &\implies \lvert g(x) - g(x_0)\rvert < \frac{\sqrt{\epsilon}}{3} + \frac{\epsilon}{3\lvert f(x_0)\rvert} && g \text{ continuous at } x_0 \end{alignat*}$$ Thus, for all $x:\lvert x-x_0\rvert < \delta=\min(\delta_f, \delta_g)$, we have $$\begin{alignat*}{2} \lvert h(x) - h(x_0)\rvert &= \lvert f(x)g(x) - f(x_0)g(x_0)\rvert &\hspace{6em}& \text{Def } h \\ &\le \lvert\{f(x) - f(x_0)\}\{g(x)-g(x_0)\}\rvert \\ &\qquad + \lvert f(x_0)\{g(x)-g(x_0)\}\rvert \\ &\qquad + \lvert g(x_0)\{f(x)-f(x_0)\}\rvert && \text{See earlier proof} \\ &< \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} && \text{\textcircled{1}}, \text{\textcircled{2}} \\ &= \epsilon \end{alignat*}$$  

Theorem. Let the function $f$ be continuous at $x_0$ and the function $g$ be continuous at $f(x_0)$. Then $h(x) = g(f(x))$ is continuous at $x_0$.

Proof. Let $\epsilon>0$. $$\begin{alignat*}{2} \text{\textcircled{1}} \qquad \exists \eta: \lvert y-f(x_0)\rvert < \eta &\implies \lvert g(y) - g(f(x_0))\rvert < \epsilon&\hspace{4em}& g \text{ continuous at } f(x_0) \\ \text{\textcircled{2}} \qquad \exists \delta: \lvert x-x_0\rvert < \delta &\implies \lvert f(x) - f(x_0)\rvert < \eta && f \text{ continuous at } x_0 \end{alignat*}$$ Thus, for all $x:\lvert x-x_0\rvert < \delta$, we have $$\begin{alignat*}{2} \lvert h(x) - h(x_0)\rvert &= \lvert g(f(x)) - g(f(x_0))\rvert &\hspace{4em}& \text{Def } h \\ &< \epsilon&& \text{\textcircled{2}} \implies \text{\textcircled{1}} \end{alignat*}$$

Exercise: Write an R function n(eps) that returns the smallest $N$ for which $n > N \implies \lvert f(x_n)-f(x_0)\rvert < \epsilon$ for $x_n = 2^{1/n}$ and $f(x) = e^x$.

Conceptually, this is a three-part process:

1. Determine what $x_n$ is converging to. Here, $x_n \to 1$.
2. Determine the largest value of delta that satisfies $e^{1+\delta} - e^1 < \epsilon$.
3. Determine the smallest value of $N$ such that $2^{1/n} - 1 < \delta$.

n <- function(eps) {
  delta <- log(eps + exp(1)) - 1
  ceiling(1/log2(1+delta))
}

Let’s test this out and make sure it works:

n(0.01)
## [1] 190
exp(2^(1/189)) - exp(1)  ## 189 not good enough
## [1] 0.01000582
exp(2^(1/190)) - exp(1)  ## 190 within 0.01
## [1] 0.009952969