Mathematics 242 – Principles of Analysis Problem Set 6 – due: Friday, March 22

’A’ Section

1. Determine whether each of the following limits exists, then prove that your answers are correct using the ε, δ definition.

(a) lim_{x→3} x^{2} − 4x + 1 Solution: The limit is −2. Proof: Given ε > 0, let δ = min(1, ε/3). For all x with 0 < |x − 3| < δ < 1, we have 2 < x < 4, so |x − 1| < 3 and hence

|x^{2} − 4x + 1 − (−2)| = |x^{2} − 4x + 3|

= |x − 3||x − 1|

< \frac{ε}{3} · 3

= ε.

This shows lim_{x→3} x^{2} − 4x + 1 = −2.

(b) lim_{x→ \frac{1}{2}} x + \frac{1}{x} Solution: The limit is \frac{5}{2}. Proof: Given ε > 0, let δ = min (\frac{1}{4}, \frac{ε}{7}). For all x with 0 < |x − \frac{1}{2} | < δ, we have

\begin{aligned}|x + \frac{1}{x} − \frac{5}{2}| & = \frac{|2x^{2} − 5x + 2|}{|2x|} \\ & = |x − \frac{1}{2}| · \frac{|x − 2|}{|x|} \end{aligned}

Since |x − \frac{1}{2} | < δ < \frac{1}{4}, we see \frac{1}{4} < x < \frac{3}{4}, so |x − 2| < \frac{7}{4} and \frac{1}{|x|} < 4. It follows that

\begin{aligned}|x − \frac{1}{2}| · \frac{|x − 2|}{|x|} & < \frac{ε}{7} · \frac{7}{4} · 4 \\ & = ε. \end{aligned}

Hence lim_{x→ \frac{1}{2}} x + \frac{1}{x} = \frac{5}{2}.

(c) lim_{x→2} \frac{x^{3}−8}{x^{2}−4}

Solution: The limit is 3. Proof: For x ≠ 2, we see

\frac{x^{3} − 8}{x^{2} − 4} = \frac{(x^{2} + 2x + 4)(x − 2)}{(x − 2)(x + 2)} = \frac{x^{2} + 2x + 4}{x + 2}

Given ϵ > 0, let δ = min(1, 3ε/4). Then for all x with 0 < |x − 2| < δ < 1, it follows that 1 < x < 3, so |x + 1| < 4 and \frac{1}{|x+2|} < \frac{1}{3}. Then

|\frac{x^{2} + 2x + 4}{x + 2} − 3| = \frac{|x^{2} − x − 2|}{|x + 2|}

= |x − 2| · \frac{|x + 1|}{|x + 2|}

< \frac{3ε}{4} · \frac{4}{3}

= ε.

This shows lim_{x→2} \frac{x^{3}−8}{x^{2}−4} = 3.

(d) Let

f(x) = \{ \begin{matrix}x sin (\frac{1}{x}) & if x ≠ 0 \\ 2 & if x = 0 \end{matrix}

and consider lim_{x→0} f(x).

Solution: The limit is 0. Proof: Given ε > 0, let δ = ε. Then for all x with 0 < |x| < δ = ε, we have

|x sin (\frac{1}{x}) − 0| = |x| |sin (\frac{1}{x})| ≤ |x| < ε

(since | sin(u)| ≤ 1 for all real u). This shows that the limit is 0 as claimed.

2. Which of the functions in question 1 are continuous at the indicated c in the limits there? Explain.

Solution: The functions in parts (a) and (b) of problem 1 are continuous at the given c since lim_{x→c} f(x) = f(c). The function in part (c) is not continuous at 2 since f(2) is not defined. The function in part (d) is not continuous at x = 0 since lim_{x→0} f(x) ≠ f(0).

3. True-False. For the true statements, give a short proof. For the false statements give a counterexample.

(a) If lim_{x→1} f(x) = e − \frac{27}{10}, then there exists a δ > 0 such that f(x) > 0 for all x with 0 < |x − 1| < δ.

This is TRUE. The reason is that e − \frac{27}{10} > 0. So if we take ε = (e − \frac{27}{10}) /2, then there is a corresponding δ > 0 such that for x with 0 < |x − 1| < δ,

|f(x) − (e − \frac{27}{10})| < (e − \frac{27}{10}) /2.

But this implies f(x) > (e − \frac{27}{10}) /2 > 0.

(b) If |f(x)| ≤ x^{3} for all x, then lim_{x→2} f(x) = 8.

Solution: This is FALSE. As a counterexample, we can take the function f(x) = 0 for all x.

(c) Let f : R → R be defined by this rule:

f(x) = \{ \begin{matrix}2x & if x is rational \\ −2x & if x is irrational. \end{matrix}

Then lim_{x→0} f(x) exists and equals 0.

Solution: This is TRUE. Given any ε > 0, let δ = ε/2. Then for all x with 0 < |x − 0| < δ = ε/2, we have |2x| = | − 2x| = 2|x| for rational and irrational x, so

|f(x)| = 2|x| < 2\frac{ε}{2} = ε.

This shows the limit is 0 as claimed.

(d) If f(x) < g(x) on a deleted neighborhood of c, lim_{x→c} f(x) = L, and lim_{x→c} g(x) = M, then L < M.

Solution: This is FALSE. Counterexample: Let f(x) = x^{4} and g(x) = x^{2}. Then f(x) < g(x) for all x with 0 < |x − 0| < 1. But lim_{x→0} f(x) = 0 = lim_{x→0} g(x). (The statement would be true if it said L ≤ M.)

’B’ Section

1. Assume that lim_{x→c} f(x) = L.

(a) Show that there exists a constant B and δ > 0 such that |f(x)| ≤ B for all x in the deleted neighborhood {x ∈ R | 0 < |x − c| < δ}.

Solution: Since lim_{x→c} f(x) = L, letting ε = 1, there is a corresponding δ > 0 such that |f(x) − L| < 1 for all x in the deleted neighborhood defined by 0 < |x − c| < δ. But for those x, L − 1 < f(x) < L + 1, so |f(x)| ≤ max(|L + 1|, |L − 1|). We can take B = max(|L + 1|, |L − 1|).

(b) Using part (a), not the limit product rule, show that lim_{x→c}(f(x))^{n} = L^{n} for all integers n ≥ 1.

Solution: Let B and δ_{0} be as in part (a). That is assume that |f(x)| ≤ B for all x with 0 < |x − c| < δ_{0}. Given ε, since lim_{x→c} f(x) = L, we have |f(x) − L| < ε/M, where

M = B^{n−1} + B^{n−2}|L| + ⋯ + B|L|^{n−2} + |L|^{n−1}

for all x with 0 < |x − c| < δ_{1} for some δ_{1} > 0. Let δ = min(δ_{0}, δ_{1}). Then for all x with 0 < |x − c| < δ, we have (using the triangle inequality on the second factor on the right side):

|(f(x))^{n} − L^{n}| = |f(x) − L||(f(x))^{n−1} + (f(x))^{n−2}L + ⋯ + f(x)L^{n−2} + L^{n−1}|

≤ |f(x) − L|(|f(x)|^{n−1} + |f(x)|^{n−2}|L| + ⋯ + |f(x)||L|^{n−2} + |L|^{n−1})

< \frac{ε}{M} (B^{n−1} + B^{n−2}|L| + ⋯ + B|L|^{n−2} + |L|^{n−1})

= \frac{ε}{M} · M = ε.

This shows lim_{x→c}(f(x))^{n} = L^{n}.

(c) Assume that f(x) ≥ 0 on some deleted neighborhood of x = c. Show that

lim_{x→c} \sqrt{f(x)} = \sqrt{L}.

(Hint: It may help to treat the cases L = 0 and L ≠ 0 separately.)

Solution: First suppose L = 0. Then for all ε > 0, there exist corresponding δ > 0 such that |f(x)| < ε^{2} for all x with 0 < |x − c| < δ. But then for the same x, we have |\sqrt{f(x)}| < ε. So lim_{x→c} \sqrt{f(x)} = 0 = \sqrt{0}. Now assume L ≠ 0 (so L > 0). Given ε > 0, there exists δ > 0 such that |f(x) − L| < ε\sqrt{L} for all x with 0 < |x − c| < δ. For these x,

|\sqrt{f(x)} − \sqrt{L}| = \frac{|f(x) − L|}{f(x) + \sqrt{L}}

≤ \frac{|f(x) − L|}{\sqrt{L}}

< \frac{ε\sqrt{L}}{\sqrt{L}}

= ε.

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

2. In this problem you will show that

lim_{θ→0} \frac{sin(θ)}{θ} = 1.

For 0 < θ < \frac{π}{2} , the point P = (cos(θ), sin(θ)) = (x, y) lies on the arc of the unit circle x^{2} + y^{2} = 1 in the first quadrant.

(a) Let O = (0, 0), Q = (cos(θ), 0), and R = (1, 0). (Draw a picture!) By considering the areas of the triangle ΔOQP and the circular sector ORP, deduce that if 0 < θ < \frac{π}{2} , then sin(θ) cos(θ) ≤ θ. (You may use “intuitively reasonable” facts about areas such as the statement that if one plane region R is completely completely contained in a second region S, then area(R) ≤ area(S).)

Solution: The area of the triangle ΔOQP is \frac{1}{2} sin(θ) cos(θ). The area of the sector is \frac{1}{2}θ, since the area of the circular sector with angle Θ of a circle of radius r is \frac{Θr^{2}}{2} . Since the sector completely contains the triangle, the desired inequality follows.

(b) Now take the tangent line to the circle at R (a vertical line), and let S = (1, tan(θ)) be the intersection of that line and the radius OP (extended). Considering the areas of the triangle ΔORS and the sector ORP as above, explain why θ ≤ tan(θ).

Solution: The area of the triangle ΔORS is \frac{1}{2} tan(θ). (By the way, in case you never have seen this before, this is the reason that the tangent function is known by that name!) This time the sector ORP is completely contained in the triangle, so the inequality follows again.

(c) Combine parts (a) and (b) to deduce that if 0 < θ < \frac{π}{2} , then

cos(θ) ≤ \frac{sin(θ)}{θ} ≤ \frac{1}{cos(θ)}.

Solution: Combining parts (a) and (b), we see

sin(θ) cos(θ) ≤ θ ≤ tan(θ) = \frac{sin(θ)}{cos(θ)}.

Since sin(θ) > 0 for the θ in this range, it follows that

cos(θ) ≤ \frac{θ}{sin(θ)} ≤ \frac{1}{cos(θ)}

and the desired inequalities follow by taking reciprocals.

(d) Using the one-sided form of Theorem 3.2.9 (The Limit Squeeze Theorem), show that

lim_{θ→0^{+}} \frac{sin(θ)}{θ} = 1.

(You will need to use the fact that cos(θ) is continuous at θ = 0.)

Solution: Since cos(θ) is continuous at θ = 0, we have lim_{θ→0^{+}} cos(θ) = 1 and hence lim_{θ→0^{+}} \frac{1}{cos(θ)} = 1 as well. By the one-sided version of the Limit Squeeze Theorem,

lim_{θ→0^{+}} \frac{sin(θ)}{θ} = 1

also.

(e) Now, for −\frac{π}{2} < θ < 0, show that \frac{sin(θ)}{θ} = \frac{sin(|θ|)}{|θ|} and use this to see that

lim_{θ→0^{−}} \frac{sin(θ)}{θ} = 1

as well.

Solution: This follows from the fact that sin is an odd function. If θ < 0, then

\frac{sin(|θ|)}{|θ|} = \frac{sin(−θ)}{−θ} = \frac{− sin(θ)}{−θ} = \frac{sin(θ)}{θ}.

By letting φ = |θ| > 0, from part (d) we see that

lim_{θ→0^{−1}} \frac{sin(θ)}{θ} = lim_{θ→0^{−}} \frac{sin(|θ|)}{|θ|} = lim_{φ→0^{+}} \frac{sin(φ)}{φ} = 1.

(f) Finally, explain how parts (d) and (e) combine to show the statement at the start of the problem.

Solution: The desired statement follows from parts (d) and (e) and Theorem 3.3.4 (equality of the two one-sided limits implies the two-sided limit exists and equals the common value).