Tag: Epsilon-Delta

Prove or disprove given statements for functions such that f(x) = o(g(x))

by RoRi

December 22, 2015

Let $f$ and $g$ be functions, both differentiable in a neighborhood of 0, with $g(x) > 0$ and such that

$f(x) = o(g(x)) \qquad \text{as} \qquad x \to 0.$

Prove or disprove each the following statements.

$\displaystyle{ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right)}$ as $x \to 0$ .
$f'(x) = o(g'(x))$ as $x \to 0$ .

True.
Proof. Since $f(x) = o(g(x))$ as $x \to 0$ we know by the definition of $o(g(x))$ that

$\lim_{x \to 0} \frac{f(x)}{g(x)} = 0.$

Thus, for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

$|x| < \delta \quad \implies \quad \left| \frac{f(x)}{g(x)} \right| < \varepsilon.$

So, for $|x| < \delta$ we have

$\begin{align*} \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| &\leq \frac{\int_0^x |f(t)| \, dt}{\left| \int_0^x g(t) \, dt \right|} \\[9pt] &< \frac{\varepsilon \int_0^x g(t) \, dt }{\left| \int_0^x g(t) \, dt \right|} \\[9pt] &= \varepsilon. \end{align*}$

The final line follows since $g > 0$ by hypothesis. Therefore,

$|x| < \delta \quad \implies \quad \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| < \varepsilon.$

Hence,

$\lim_{x \to 0} \frac{ \int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} = 0.$

By definition, we then have

$\int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right). \qquad \blacksquare$
False.
Consider $f(x) = x^2 \sin \left( \frac{1}{x} \right)$ for $x \neq 0$ and $f(x) = 0$ for $x = 0$ . Then, for $x \neq 0$ ,

$f'(x) = 2x \sin \left( \frac{1}{x} \right) - \cos \left( \frac{1}{x} \right).$

For $x = 0$ we have $f'(0) = 0$ .

Next,

$\begin{align*} f(x) &= x^2 \sin \left( \frac{1}{x} \right) \\[9pt] &= x^2 \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right) \left( \frac{1}{x} \right) \\[9pt] &= x \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right). \end{align*}$

Since $\lim_{x \to 0} \frac{\sin (1/x)}{1/x} = 1$ we have $f(x) = o(x)$ as $x \to 0$ . However, $f'(x) \neq o(1)$ since

$\lim_{x \to 0} \left( 2 \sin \left( \frac{1}{x}\right) - \cos \left( \frac{1}{x} \right) \right)$

does not exist.

Prove that sin (1/x) has no limit as x approaches 0

by RoRi

August 29, 2015

Define the function:

$f(x) = \sin \frac{1}{x} \qquad \text{for } x \neq 0.$

Prove there is no $A \in \mathbb{R}$ such that

$\lim_{x \to 0} f(x) = A.$

Proof. We prove this by contradiction. Suppose there does exist some $A \in \mathbb{R}$ such that $\lim_{x \to 0} f(x) = A$ . From the definition of limit this means that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that

$| f(x) - A | < \varepsilon \qquad \text{whenever} \qquad 0 < |x| < \delta.$

First, we claim that for any such $A$ we must have $|A| \leq 1$ . This must be the case since if $|A| > 1$ then $|A| - 1 > 0$ , so we may choose $\varepsilon$ such that $0 < \varepsilon < (|A|-1)$ . But then, since $|f(x)| \leq 1$ for all $x$ , we have

$|f(x) - A| = |A - f(x)| \geq |A| - |f(x)| \geq |A| - 1 > \varepsilon.$

(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality $|A - f(x) \geq |A| - |f(x)|$ .) This contradicts our choice of $\varepsilon$ , so $|A|$ must be less than or equal to 1.
Next, suppose $|A| \leq 1$ . Then choose $\varepsilon = \frac{1}{2} > 0$ . To obtain our contradiction we must show that there is no $\delta > 0$ such that

$|f(x) - A| < \frac{1}{2}, \qquad \text{whenever} \qquad 0 < |x| < \delta.$

By the Archimedean property of the real numbers we know that for any $x \in \mathbb{R}$ , there exists a positive integer $n$ with $n \equiv 1 \pmod{4}$ such that $0 < \frac{2}{n \pi} < |x|$ . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being $1 \pmod{4}$ .) But,

$0 < \frac{2}{n \pi} < |x| \quad \implies \quad 0 < \frac{2}{(n+2) \pi} < |x|.$

Then, from the definition of $f$ and since $n \equiv 1 \pmod{4}$ and $n+2 \equiv 3 \pmod{4}$ we have,

$\begin{align*} f \left( \frac{2}{n \pi} \right) &= \sin \left( \frac{n \pi}{2} \right) = 1 \\ f \left( \frac{2}{(n+2) \pi} \right) &= \sin \left( \frac{(n+2)\pi}{2} \right) = -1. \end{align*}$

But then,

$\begin{align*} \left| f \left( \frac{n \pi}{2} \right) - A \right| < \frac{1}{2} && \implies && |1 - A| &< \frac{1}{2} \\ && \implies && \frac{1}{2} &< A \leq 1 &(\text{from above } A \leq 1). \end{align*}$

Furthermore,

$\left| f \left( \frac{(n+2) \pi}{2} \right) - A \right| = |-1-A| > \frac{1}{2}.$

This contradicts that if $\lim_{x \to 0} f(x) = A$ then for every $\varepsilon > 0$ we have $|f(x) - A| < \varepsilon$ for all $0 < |x| < \delta$ . In other words, we found an $\varepsilon$ greater than 0 (in particular, we found $\varepsilon = \frac{1}{2}$ ) such that no matter how small we choose $\delta$ there exists an $x$ such that $|x|$ is smaller than $\delta$ , but $|f(x) - A|$ is bigger than $\varepsilon$ . This exactly contradicts the definition of $\lim_{x \to 0} f(x) = A$ . Hence, there can be no such number $A \in \mathbb{R}. \qquad \blacksquare$

This proof shows that there is now way to define $f(0)$ so that $f$ is continuous at 0. We see this since for $f$ to be continuous at 0 we must have $\lim_{x \to 0} f(x) = f(0)$ . But if we try to define $f(0) = A$ for any $A \in \mathbb{R}$ , we will fail to achieve continuity since the proof shows $\lim_{x \to 0} f(x) \neq A$ for any $A$ we might want to choose.

Give an alternate proof of the continuity of the sine and cosine functions

by RoRi

August 29, 2015

Use the inequality
$| \sin x | < |x|, \qquad \text{for } 0 < |x| < \frac{\pi}{2},$

to prove that the sine function is continuous at 0.
Recall the trig identity,
$\cos 2x = 1 - 2 \sin^2 x.$

Use this and part (a) to prove that the cosine function is continuous at 0.
Use the formulas for sine and cosine of a sum to prove that the sine and cosine functions are continuous for all $x \in \mathbb{R}$ .

Proof. To show $\sin$ is continuous at 0, we must show that $\lim_{x \to 0} \sin x = \sin 0 = 0$ . We show this limit is zero directly from the epsilon-delta definition of the limit, i.e., given arbitrary positive $\varepsilon$ , we have $|\sin x| < \varepsilon$ whenever $|x| < \delta$ . Let $\varepsilon$ be an arbitrary number greater than 0, $\varepsilon > 0$ . Then, let $\delta = \varepsilon$ . Using the given inequality we have,
$| \sin x | < |x| < \delta = \varepsilon, \qquad \text{whenver } 0 < |x| < \delta.$

Thus, $\sin x$ is continuous at $0. \qquad \blacksquare$
Proof. First, using the given trig identity we have
$\cos (2x) = 1 - 2 \sin^2 x \quad \implies \quad \cos x = 1 - 2 \sin^2 \frac{x}{2}.$

Thus,

$\begin{align*} \lim_{x \to 0} \cos x &= \lim_{x \to 0} \left( 1 - 2 \sin^2 \frac{x}{2} \right) \\ &= 1 - 2 \cdot \lim_{x \to 0} \left( \sin \left( \frac{x}{2} \right) \sin \left( \frac{x}{2} \right) \right) \\ &= 1 = \cos 0. \end{align*}$

Thus, cosine is continuous at 0.
Proof. Finally, to show sine and cosine are continuous for all $x \in \mathbb{R}$ , we show that $\lim_{x \to h} \sin x = sin h$ , and $\lim_{x \to h} \cos x = cos h$ . First, we recall the formulas for the sine and cosine of a sum,
$\begin{align*} \sin (x+h) &= \sin x \cos h + \sin h \cos x \\ \cos (x+h) &= \cos x \cos h - \sin x \sin h. \end{align*}$

So, we compute the limits

$\begin{align*} \lim_{x \to h} \sin x &= \lim_{x \to 0} \sin (x+h) = \lim_{x \to 0} (\sin x \cos h + \sin h \cos x) = \sin h \\ \lim_{x \to h} \cos x &= \lim_{x \to 0} \cos (x+h) = \lim_{x \to 0} (\cos x \cos h - \sin x \sin ) = \cos h. \end{align*}$

Therefore, sine and cosine are continuous for all $x \in \mathbb{R}. \qquad \blacksquare$