Mathematics 242 – Principles of Analysis Selected Solutions, Discussion 4 October 4, 2000

C 4) Given: z_{n} = \frac{2+(−1)^{n}}{n}, a sequence converging to zero as n → ∞. Note the terms of this sequence and their values in f(x) start out like this:

\begin{matrix}n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & ⋯ \\ z_{n} & \frac{1}{1} & \frac{3}{2} & \frac{1}{3} & \frac{3}{4} & \frac{1}{5} & \frac{3}{6} = \frac{1}{2} & \frac{1}{7} & \frac{3}{8} & \frac{1}{9} & \frac{3}{10} & ⋯ \\ f(z_{n}) & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{2} & \frac{1}{5} & \frac{1}{2} & \frac{1}{7} & \frac{1}{2} & \frac{1}{3} & \frac{1}{3} & ⋯ \end{matrix}

The thing you should notice about this is that there does not seem to be any overall pattern to the values f(z_{n}).

Indeed, let’s consider the following subsequences. Let n_{k} be the kth prime number bigger than 2. So n_{1} = 3, n_{2} = 5, n_{3} = 7, n_{4} = 11, etc. All these are odd, so z_{n_{k}} = \frac{1}{n_{k}}. To compute the value of the function f on these, we need the second alternative, since the denominator is the first power of a prime. f(z_{n_{k}}) = z_{n_{k}} and this subsequence of the sequence f(z_{n}) converges to zero.

On the other hand, consider a second subsequence z_{m_{k}} where m_{k} is the kth even number that is not a power of 2, and is not divisible by 3. So m_{1} = 10, m_{2} = 14, n_{3} = 20, n_{4} = 22, and so forth. For all of the m_{k}, we have that z_{m_{k}} is already in lowest terms and equals 3 divided by an even number that is not a power of 2. Hence the last alternative in the definition of f applies to all of the z_{m_{k}}, and f(z_{m_{k}}) = \frac{1}{3} for all k.

Since f(z_{n}) has subsequences converging to different values, it cannot be true that lim_{n→∞} f(z_{n}) exists. Since z_{n} is a legal sequence converging to zero (that is, a “way for x to approach zero” with x ≠ 0) this says that lim_{x→0} f(x) cannot exist either.

D) I was looking for something like the following: In order to use sequences x_{n} → a to decide whether lim_{x→a} f(x) = L, we would have to be able to look at every possible sequence x_{n} → a (but with x_{n} ≠ a for all n) and decide whether lim_{n→∞} f(x_{n}) = L. Let’s call those sequences x_{n} → a with x_{n} ≠ a for all n “legal” sequences. In Question A we have three different “legal” sequences converging to 3, and f(x_{n}) is converging to 11 for all 3. This suggests that lim_{x→3} f(x) = 11, but it is not a complete proof – we have not checked that lim_{x→3} f(x) = 11 for x → 3 in all possible ways. In Question B, the first two sequences are “legal” sequences converging to zero, and lim_{n→∞} f(x_{n}) = 0 for each. This again suggests that lim_{x→0} f(x) = 0. The sequence from part 4 is not “legal” though, since some of its terms equal zero (all the odd-numbered terms). Therefore, the fact that f(z_{n}) does not converge does not contradict our expectation that lim_{x→0} f(x) = 0 (!) Finally, in question C, we see that there is a “legal” sequence x_{n} converging to zero such that f(x_{n}) converges to 0, another “legal” sequence y_{n} where f(y_{n}) converges to \frac{1}{7}, and a third “legal” sequence z_{n} where f(z_{n}) does not converge at all. Hence lim_{x→0} f(x) cannot exist. (The values f(x) are not approaching any one limiting value as x → 0.)