Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 3 October 20, 2000

Section 3.1

3. We are given that |f(x)| ≤ 7x^{2} for all x. This is equivalent to the inequalities

(1)  −7x^{2} ≤ f(x) ≤ 7x^{2}

for all x.

a. Since lim_{x→0} −7x^{2} = 0 = lim_{x→0} 7x^{2}, the Limit Squeeze Theorem shows that the

limit of f(x) is lim_{x→0} f(x) = 0. b. Here we have lim_{x→0} −7x^{2} = −28 but lim_{x→0} 7x^{2} = 28. The Squeeze Theorem does

not apply, and we cannot say more than the following: If the limit exists, then by Theorem 3.2.4, it is between -28 and +28. However, there is no guarantee the limit exists at all in this case. f(x) could have a jump discontinuity (or worse) at x = 2.

9. We are given that lim_{x→c} f(x) exists. Hence for all ϵ > 0, there exist δ > 0 such that |f(x)−L| < ϵ for all x with 0 < |x−c| < δ. With ϵ = 1, for instance, we get a corresponding δ such that |f(x) − L| < 1 for all x in the deleted neighborhood 0 < |x − c| < δ of c. Here’s a clever way to get an upper bound on |f(x)| using the triangle inequality (idea taken from Jackie Pournaras’s solution): For x in this deleted neighborhood of c:

|f(x)| = |f(x) − L + L| ≤ |f(x) − L| + |L| < 1 + |L|.

Hence M = 1 + |L| is one upper bound.

10. We are given that lim_{x→c} f(x) = L a. For all ϵ > 0, there exist δ > 0 such that |f(x) − L| < ϵ for all x with 0 < |x − c| < δ.

But then for x in the same deleted neighborhood of c, by a general property of absolute values (Exercise 4, Section 1.3):

||f(x)| − |L|| ≤ |f(x) − L| < ϵ.

Hence lim_{x→c} |f(x)| = |L|. b. Factoring the difference of squares, we have |f^{2}(x) − L^{2}| = |f(x) + L| · |f(x) − L|.

By problem 9, we know that there is some δ_{1} and M > 0 such that |f(x)| ≤ M for all x with 0 < |x − c| < δ_{1}. Hence, by the triangle inequality, |f(x) + L| ≤ |f(x)| + |L| ≤ M + |L| for all x in the deleted neighborhood 0 < |x − c| < δ_{1}. Then since lim_{x→c} f(x) = L, for any ϵ > 0, there exists a δ_{2} such that |f(x) − L| < \frac{ϵ}{M+|L|} for all x with 0 < |x − c| < δ_{2}. But then with δ = min(δ_{1}, δ_{2}), we get for all x with 0 < |x − c| < δ:

|f^{2}(x) − L^{2}| = |f(x) + L| · |f(x) − L|

≤ (M + |L|) · \frac{ϵ}{M + |L|}

= ϵ

Hence lim_{x→c} f^{2}(x) = L^{2}.

Section 3.2 7. Given that f(x) ≥ 0 on a deleted neighborhood of c, and lim_{x→c} f(x) = L.

a) Suppose L = 0. Then for all ϵ > 0, there exists a δ > 0 such that |f(x)| < ϵ^{2} for all x with 0 < |x − c| < δ. But then, for x in this deleted neighborhood of c,

|\sqrt{f(x)}| < \sqrt{ϵ^{2}} = ϵ,

which shows that lim_{x→c} \sqrt{f(x)} = 0 = \sqrt{0}.

b) Now suppose that L > 0. We have

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

where the last inequality follows since \sqrt{f(x)} ≥ 0 for all x in the deleted neighborhood of c. Then since lim_{x→c} f(x) = L, for any ϵ > 0, there exists a δ such that |f(x)−L| < ϵ\sqrt{L} for all x with 0 < |x − c| < δ. For x in this deleted neighborhood,

|\sqrt{f(x)} − \sqrt{L}| ≤ \frac{|f(x) − L|}{\sqrt{L}} < \frac{ϵ\sqrt{L}}{\sqrt{L}} = ϵ

Therefore, lim_{x→c} \sqrt{f(x)} = \sqrt{L}.

8. The idea here is to use the limit theorems from this section, not the ϵ - δ definition of limits. Given lim_{x→c} f(x) = L a. To show that lim_{x→c} |f(x)| = |L|, consider the cases L > 0, L = 0, L < 0. First,

if L > 0, then Theorem 3.2.2 says that there is a deleted neighborhood of c where f(x) > 0. We only need to consider this neighborhood to compute lim_{x→c} |f(x)| = |L|. But on this neighborhood, |f(x)| = f(x) and |L| = L, so there is really nothing to prove. Next, if L < 0, then by the same theorem, there is a deleted neighborhood of c where f(x) < 0, and we only need consider x in that neighborhood to compute the limit. We have by the limit product rule (Theorem 3.2.3) with one function the constant −1:

lim_{x→c} |f(x)| = lim_{x→c} −f(x)

= (lim_{x→c} −1)(lim_{x→c} f(x))

= (−1)(L) = |L|

Finally, if L = 0, then |f(x)| must also be going to 0 as x → c (why?) b. Use the limit product rule (Theorem 3.2.3 b) with g(x) = f(x):

lim_{x→c} f(x)^{2} = (lim_{x→c} f(x))(lim_{x→c} f(x)) = L · L = L^{2}

Section 3.3

5 c) We must show that lim_{x→4^{−}} \frac{2x+3}{x−4} = −∞. By the definition, this means that we must show that for all M > 0 there exists a δ > 0 such that

(2)  \frac{2x + 3}{x − 4} < −M

for all x in some half-neighborhood 4 − δ < x < 4. Solving (2) for x, if x < 4, so x − 4 < 0 we get

\frac{2x + 3}{x − 4} < −M ⇔ 2x + 3 > −(x − 4)M

⇔ 2x + 3 > −Mx + 4M ⇔ (M + 2)x > 4M − 3

⇔ x > \frac{4M − 3}{M + 2}

⇔ x > \frac{4M + 8 − 11}{M + 2}

⇔ x > 4 − \frac{11}{M + 2}

Hence if we take δ = \frac{11}{M+2} > 0, then 4 − δ < x < 4 implies \frac{2x+3}{x−4} < −M, which is what we had to show.