Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 6 November 8, 2004

20.4 c) We want to show that lim_{x→2} x^{3} = 8. Note that |x^{3} − 8| = |x − 2| · |x^{2} + 2x + 4|. If |x − 2| < 1, for instance, then 1 < x < 3, so 7 < x^{2} + 2x + 4 < 19. So we are ready for the proof:

Let ε > 0, and let δ = min(1, ε/19). Then for all x with 0 < |x − 2| < δ, we have

|x^{3} − 8| = |x − 2| · |x^{2} + 2x + 4| < ε/19 · 19 = ε.

Hence lim_{x→2} x^{3} = 8.

20.6 b) Given: f(x) ≥ 0 for all x ∈ D and lim_{x→c} f(x) = L. Note first that since \sqrt{f(x)} ≥ 0,

|\frac{1}{\sqrt{f(x)} + \sqrt{L}}| < \frac{1}{\sqrt{L}},

for all x ∈ D. Now let ε > 0 be arbitrary. Since lim_{x→c} f(x) = L, there exists a δ > 0 such that |f(x) − L| < ε · \sqrt{L} for all x ∈ D with |x − c| < δ. But then if 0 < |x − c| < δ, we have by the difference of squares factorization:

|\sqrt{f(x)} − \sqrt{L}| = |f(x) − L| · |\frac{1}{\sqrt{f(x)} + \sqrt{L}}| < ε · \sqrt{L} · \frac{1}{\sqrt{L}} = ε.

This shows lim_{x→c} \sqrt{f(x)} = \sqrt{L}.

20.7 b) The function f(x) = sin(1/x) has no limit as x → 0^{+} because there exist sequences s_{n} converging to zero for which the sequences of values f(x) have different limits. For instance, with s_{n} = \frac{1}{nπ} we have f(s_{n}) = sin(nπ) = 0 all n. Hence lim_{n→∞} f(s_{n}) = 0. But for t_{n} = \frac{2}{(4n+1)π}, we have sin(t_{n}) = sin(2nπ + π/2) = 1 for all n. Hence lim_{n→∞} f(t_{n}) = 1. By Theorem 20.8, lim_{x→0^{+}} f(x) does not exist.

c) lim_{x→0^{+}} x sin(1/x) = 0. Proof: Let ε > 0, and take δ = ε. Then for all x > 0 with 0 < |x − 0| < δ = ε,

|x sin(1/x)| = |x|| sin(1/x)| ≤ |x| · 1 < |x| < ε.

(We are using the fact that | sin(u)| ≤ 1 for all u here.)

20.8 a) We want to show that if lim_{x→c} f(x) = L, then the limit L is unique, first using the definition. Suppose lim_{x→c} f(x) = L and lim_{x→c} f(x) = M are both true. Then for all ε > 0, there exist δ_{1} > 0 such that |f(x) − L| < ε/2 for all x ∈ D with 0 < |x − c| < δ_{1} and similarly there exists δ_{2} such that |f(x) − M| < ε/2 for all x ∈ D with 0 < |x − c| < δ_{2}. Then for δ = min(δ_{1}, δ_{2}) and all x ∈ D with 0 < |x − c| < δ

|L − M| = |L − f(x) + f(x) − M| ≤ |f(x) − L| + |f(x) − M| < ε/2 + ε/2 = ε

Since 0 ≤ |L − M| < ε for all positive ε, we must have |L − M| = 0, so L = M.

b) Now we prove the same result a different way, using the “sequential criterion” for limits (Theorem 20.8), and the uniqueness for limits of sequences that we proved earlier. We assume that lim_{x→c} f(x) = L and lim_{x→c} f(x) = L as in part a. Then by Theorem 20.8, for any sequence s_{n} in D with s_{n} ≠ c for all n and lim_{n→∞} s_{n} = c, we have that f(s_{n}) is convergent to the limit of f(x). Thus we have have lim_{n→∞} f(s_{n}) = L and lim_{n→∞} f(s_{n}) = M. But Theorem 16.14, proved before, says: The limit of any convergent sequence is unique. So L = M.

20.11. The idea for this one is like the proof in 8 b) above. We assume that f(x) ≤ g(x) ≤ h(x) for all x ∈ D, and that lim_{x→c} f(x) = L = lim_{x→c} h(x). Then Theorem 20.8 says: If s_{n} is any sequence in D converging to c (with s_{n} ≠ c for all n), we have lim_{n→∞} f(s_{n}) = lim_{n→∞} h(s_{n}) = L. But now we use the result from Theorem 17.4 (proved previously): If t_{n} ≤ u_{n} for all n, and both sequences are convergent, then lim_{n→∞} t_{n} ≤ lim_{n→∞} u_{n}. As in Exercise 17.20, since f(s_{n}) ≤ g(s_{n}) ≤ h(s_{n}) for all n, we have that g(s_{n}) must be convergent to L as well. (Proof: Since lim_{n→∞} f(s_{n}) = lim_{n→∞} h(s_{n}), for all ε > 0, there exists N such that |f(s_{n}) − L| < ε and |h(s_{n}) − L| < ε for all n ≥ N. This says L − ε < f(s_{n}) ≤ g(s_{n}) ≤ h(s_{n}) < L + ε for all n ≥ N. So |g(s_{n}) − L| < ε also. Since this is true for all sequences in D converging to c, lim_{x→c} g(x) = L by Theorem 20.8.