# Recitation 2

Poem: Epsilon and Delta: A Little Love Story by Bonnie Shulman (1994).

Definition Suppose $f\colon A\to B$ is a function, where $A$ and $B$ are subsets of $\mathbb{R}$. We say $\lim_{x\to x_0} f(x) = L$ if for every $\epsilon > 0$ there exists $\delta > 0$ such that

$$|f(x) - L| < \epsilon \text{ for all }0<|x-x_0|<\delta.$$

Problem 1: If $\lim_{x\to 0}f(x) = 0$, then also $\lim_{x\to 0}f(5x) = 0$.

Sketch: For every $\epsilon > 0$, there is $\delta_1 > 0$ such that $|f(x)| < \epsilon$ for all $0<|x|<\delta_1$. Now take $\delta_2 = \delta_1/5$ (why?). Check $|f(5x)|<\epsilon$ for all $0<|x|<\delta_2$.

Problem 2: Prove that $\lim_{x\to x_0}\sin x = \sin x_0$ and $\lim_{x\to x_0} \cos x = \cos x_0$.

Idea: Use $|\sin x - \sin x_0| \le |x-x_0|$ and $|\cos x - \cos x_0|\le|x-x_0|$ from the homework.

Fact 1: If $\lim_{x\to x_0}f(x) = L$ and $L > 0$, then $\lim_{x\to x_0}\sqrt{f(x)} = \sqrt{L}$.

Fact 2: If $f(x) = g(x)$, then $\lim_{x\to x_0}f(x)$ exists if and only if $\lim_{x\to x_0} g(x)$ exists. Moreover, when this happens, the limits are equal.

Problem 3: We know that $\forall x\ f(2x) = \sqrt{f(x) + 5}$ and that $\lim_{x\to 0} f(x) = L$. Find $L$.

Sketch: Step 1: show that $\lim_{x\to 0} f(2x) = L$ (compare with problem 1). Step 2.1: show that $\lim_{x\to 0}f(x) + 5 = L + 5$. Step 2.2: use Fact 1 show that $\lim_{x\to 0}\sqrt{f(x) + 5} = \sqrt{L + 5}$ (why $L + 5 > 0$?). Using Fact 2 and results from Step 1 and 2, we get $L = \sqrt{L+5}$. Solve quadratic equation and pick the positive solution (why is the negative solution discarded?).

Riddle: Give an example of $f, g$ such that $\lim_{x\to 0}f(x) = 0, \lim_{x\to 0} g(x) = 0$ and $\lim_{x\to 0} f(g(x)) = 0$.

Problem 4: Let $f(x) = \sin(1/x)$. Show that $\lim_{x\to 0} f(x)$ does not exist.

Fact 3: Suppose $\lim_{x\to x_0}f(x)$ exists. For every $\epsilon > 0$, there exists $\delta > 0$ such that $|f(x_1) - f(x_2)|<\epsilon$ for all $0<|x_1-x_0|, |x_2 - x_0|<\delta$.

Proof strategy: The contraposition of Fact 3 gives a way to show that the limit does not exist. If there exists $\epsilon > 0$ such that for every $\delta > 0$, $|f(x_1)-f(x_2)|\ge \epsilon$ for some $0<|x_1 - x_0|,|x_2 - x_0| < \delta$, then $\lim_{x\to x_0}f(x)$ does not exist.

Proof of Problem 4: Take $\epsilon = 1$. For every $\delta > 0$, pick a natural number $n$ such that $x_1 := \frac{1}{n\pi - \pi/2}$ and $x_2 := \frac{1}{n\pi + \pi/2}$ satisfies $0<|x_1|,|x_2|<\delta$ (why can we find such $n$?). Notice that $1/x_1 = n\pi - \pi/2$ and $1/x_2 = n\pi + \pi/2$. Thus $|f(x_1)-f(x_2)|=2>1=\epsilon$. By the contraposition of Fact 3, $\lim_{x\to x_0}f(x)$ does not exist.

Problem 5: Show that if $g$ is bounded and $\lim_{x\to 0} f(x) = 0$, then $\lim_{x\to 0} f(x)g(x) = 0$.

Proof: Because $g$ is bounded, by definition, there is $M > 0$ such that $-M \le g(x) \le M$. For every $\epsilon > 0$, since $\lim_{x\to 0} f(x) = 0$, there is $\delta > 0$ such that $|f(x)| < \epsilon/M$ for all $0<|x|<\delta$ (why can we choose $\epsilon/M$?), thus $|f(x)g(x)|\le \epsilon/M \cdot M = \epsilon$. By the definition of limit, $\lim_{x\to 0}f(x)g(x) = 0$.

Application of Problem 5: Show that $\lim_{x\to 0}x\sin(1/x) = 0$.