Solutions to Problem Set 9

Math 310-01/02

9-25-2017

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

$$\lim_{x\to 3} (4 x + 5) = 17.$$

Let $\epsilon > 0$ . Set $\delta = \dfrac{\epsilon}{4}$ . Then

$$\eqalign{\dfrac{\epsilon}{4} = \delta &> |x - 3| \cr \epsilon &> |4 x - 12| \cr \epsilon &> |(4 x + 5) - 17| \cr}$$

Hence, $\displaystyle \lim_{x\to
   3} (4 x + 5) = 17$ .


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

$$\lim_{x \to 3} (2 x^2 + x) = 21.$$

Let $\epsilon > 0$ . Set $\delta = \min \left(1, \dfrac{\epsilon}{15}\right)$ . Then

$$1 \ge \delta \quad\hbox{and}\quad \dfrac{\epsilon}{15} \ge \delta.$$

First,

$$1 \ge \delta > |x - 3|.$$

So

$$\eqalign{ 2 & < x < 4 \cr 4 & < 2 x < 8 \cr 11 & < 2 x + 7 < 15 \cr}$$

Therefore,

$$15 > |2 x + 7|.$$

Also,

$$\dfrac{\epsilon}{15} \ge \delta > |x - 3|.$$

Multiplying the last two inequalities, I get

$$\epsilon = 15 \cdot \dfrac{\epsilon}{15} > |2 x + 7| |x - 3| = |2 x^2 + x - 21|.$$

This proves that $\displaystyle
   \lim_{x \to 3} (2 x^2 + x) = 21$ .


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