QUIZ #2: CALCULUS 1A (Stankova) Wednesday, February 4, 2004 Section 10:00–11:00 (Voight)

Problem 1. Evaluate the limit, if it exists:

lim_{h→0} \frac{(3 + h)^{−1} − 3^{−1}}{h} .

Solution. Applying limit laws, we get the indeterminate expression 0/0. Thus we try to simplify and factor:

lim_{h→0} \frac{(3 + h)^{−1} − 3^{−1}}{h} = lim_{h→0} \frac{\frac{1}{3 + h} − \frac{1}{3}}{h}

= lim_{h→0} \frac{1}{h} (\frac{1}{3 + h} − \frac{1}{3})

= lim_{h→0} \frac{1}{h} (\frac{3 − (3 + h)}{3(3 + h)})

= lim_{h→0} \frac{−h}{3h(3 + h)} = lim_{h→0} \frac{−1}{3(3 + h)} \begin{matrix}LL \\ = \end{matrix} − 1/9.

QUIZ #2: CALCULUS 1A (Stankova) Wednesday, February 4, 2004 Section 11:00–12:00 (Voight)

Problem 1. Use the given graph of f to find a number δ such that

|f(x) − 3| < 0.3 whenever 0 < |x − 4| < δ.

Solution. Reading directly from the graph, we see that x can deviate to the left by 4 − 2.89 = 1.11 and to the right by 5.29 − 4 = 1.29. We can only deviate to the left or right by the smaller of these values, so we may take δ = 1.11.

Problem 2. Prove the statement using the ϵ, δ definition of limit. Illustrate with a graph.

lim_{x→a} c = c.

[Hint: There may be more than one correct answer. Justify your reasoning.]

Solution. We claim that lim_{x→a} c = c. Let ϵ > 0 be given. Then we need to show that |f(x) − L| < ϵ whenever |x − a| < δ. But the first inequality is just |c − c| = 0 < ϵ, which is always true. Therefore we may choose δ to be any real number.

[If the space aliens came, we would tell them that their inequality is always satisfied, and give them any value of δ no matter what value of ϵ they gave us. Suckers!]