Homework 8

Solution.

1. Let $x, y \in \R$. By the mean value theorem, there exists $c$ between $x$ and $y$ such that

   $$\sin{x} - \sin{y} = \p{x - y} \cos{c}.$$

   But $\abs{\cos{\theta}} \leq \theta$ for any $\theta \in \R$, so taking absolute values, we get

   $$\abs{\sin{x} - \sin{y}} = \abs{\p{x - y} \cos{c}} \leq \abs{x - y}.$$

2. Given $\varepsilon > 0$, we can just set $\delta = \varepsilon$. If $\abs{x - y} < \delta$, then by (1), we have

   $$\abs{\sin{x} - \sin{y}} \leq \abs{x - y} < \delta = \varepsilon.$$

Solution.

1. By the product rule and chain rule,

   $$f'\p{a} = 2a \sin\frac{1}{a} - \cos\frac{1}{a}.$$

2. We have

   $$\abs{\frac{f\p{x} - f\p{0}}{x - 0}} = \abs{\frac{x^2\sin\frac{1}{x}}{x}} = \abs{x\sin \frac{1}{x}} \leq \abs{x}.$$

   By the squeeze theorem (which still applies to limits of functions by using the sequential definition of limits), since $\abs{x} \to 0$ as $x \to 0$, we see that $f$ is differentiable at $0$ and $f'\p{0} = 0$.

3. Let $x_n = \frac{1}{2\pi n}$ so that $x_n \to 0$. Then

   $$f'\p{x_n} = \frac{1}{\pi n} \sin\p{2\pi n} - \cos\p{2\pi n} = 1$$

   for all $n$. Thus, $f'\p{x_n} \to 1$, but $f'\p{0} = 0$.

Solution.

"$\implies$"

Assume $f'\p{a}$ exists; that is, the limit

exists. Let

(You can figure this out by just solving for $\varepsilon\p{x}$ in the equation they give you. Also, it doesn't matter what the value of $\varepsilon\p{a}$ is as long as you define it.) Then it's easy to check that

so it remains to check that $\lim_{x\to a} \varepsilon\p{x} = 0$. By limit laws, it exists and

"$\impliedby$"

Suppose there exist a number $L \in \R$ and a function $\varepsilon\p{x}$ such that

By limit laws again $f$ is differentiable at $a$ with

Common Mistakes

The most common mistake I saw was trying to do something like this to try to show that $\lim_{x\to a} \varepsilon\p{x} = 0$:

$$f'\p{a} = \lim_{x\to a} \br{\frac{f\p{x} - f\p{a}}{x - a}} = \lim_{x\to a} \br{f'\p{a} - \varepsilon\p{x}} = f'\p{a} - \lim_{x\to a} \varepsilon\p{x}.$$

Thus, solving gives $\lim_{x\to a} \varepsilon\p{x} = 0$.

The issue with this is that the argument only says this: if $\lim_{x\to a} \varepsilon\p{x}$ exists, then it must be $0$. However, you need to show that $\lim_{x\to a} \varepsilon\p{x}$ exists to begin with. In other words, you can't just take limits on both sides and solve. The easiest way to do this is to use limit laws, which tells you that the limit exists and tells you have to compute the limit--you kill two birds with one stone.

This argument can be fixed, though it requires a little bit of work. See the following exercise.