# 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.