Solutions to Problem Set 11

Math 310-01/02

10-6-2017

1. Use the $\epsilon-\delta$ definition of a limit to prove that

$$\lim_{x \to 2} \dfrac{2 x + 5}{x + 1} = 3.$$

Let $\epsilon > 0$ . Set $\delta = \min\left(1, 2\epsilon\right)$ , and assume $\delta > |x - 2|$ . Then

$$1 \ge \delta \quad\hbox{and}\quad 2\epsilon \ge \delta.$$

First,

$$\eqalign{ 1 \ge \delta & > |x - 2| \cr 1 < x & < 3 \cr 2 < x + 1 & < 4 \cr \noalign{\vskip2pt} \dfrac{1}{2} > \dfrac{1}{x + 1} & > \dfrac{1}{4} \cr}$$

Hence, $\dfrac{1}{2} >
   \left|\dfrac{1}{x + 1}\right|$ .

In addition,

$$2\epsilon \ge \delta > |x - 2|.$$

Multiplying the last two inequalities,

$$\eqalign{ \epsilon = \dfrac{1}{2} \cdot 2\epsilon & > \left|\dfrac{1}{x + 1}\right||x - 2| \cr & = \left|\dfrac{x - 2}{x + 1}\right| \cr & = \left|\dfrac{2 - x}{x + 1}\right| \cr & = \left|\dfrac{(2 x + 5) - 3(x + 1)}{x + 1}\right| \cr & = \left|\dfrac{2 x + 5}{x + 1} - 3\right| \cr}$$

This proves that $\displaystyle
   \lim_{x \to 2} \dfrac{2 x + 5}{x + 1} = 3$ .


2. Premises: $\displaystyle
   \left\{\matrix{C \ifthen A \cr \lnot B \ifthen \lnot D \cr D \lor C
   \cr}\right.$ .

Prove: $A \lor B$ .

$$\matrix{ \hfill 1. & C \ifthen A \hfill & \hbox{Premise} \hfill \cr \hfill 2. & \lnot B \ifthen \lnot D \hfill & \hbox{Premise} \hfill \cr \hfill 3. & D \lor C \hfill & \hbox{Premise} \hfill \cr \hfill 4. & B \hfill & \hbox{Proof by cases - Case 1} \hfill \cr \hfill 5. & A \lor B \hfill & \hbox{Constructing a disjunction (4)} \hfill \cr \hfill 6. & \lnot B \hfill & \hbox{Proof by cases - Case 2} \hfill \cr \hfill 7. & \lnot D \hfill & \hbox{Modus ponens (2,6)} \hfill \cr \hfill 8. & C \hfill & \hbox{Disjunctive syllogism (3,7)} \hfill \cr \hfill 9. & A \hfill & \hbox{Modus ponens (1,8)} \hfill \cr \hfill 10. & A \lor B \hfill & \hbox{Constructing a disjunction (9)} \hfill \cr \hfill 11. & A \lor B \hfill & \hbox{Proof by cases (4,5,6,10)} \quad\halmos \hfill \cr}$$


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