Question

the function f in the given figure satisfies lim x→2 f(x)=6. for each value of ε, find all values of δ>0 such that |f(x) - 6|<ε whenever 0<|x - 2|<δ. (a) ε = 2 (b) ε = 1 (c) for any ε, make a conjecture about the corresponding value of δ satisfying the above description. (a) when ε = 2, the values of δ>0 such that |f(x) - 6|<ε whenever 0<|x - 2|<δ satisfy the inequality 0<δ≤1. (type an inequality using δ as the variable. simplify your answer.) (b) when ε = 1, the values of δ>0 such that |f(x) - 6|<ε whenever 0<|x - 2|<δ satisfy the inequality 0<δ≤1/2. (type an inequality using δ as the variable. simplify your answer.) (c) from parts (a) and (b), for any ε>0, make a conjecture that the values of δ>0 such that |f(x) - 6|<ε whenever 0<|x - 2|<δ satisfy the inequality δ = ε/2. (type an inequality using δ and ε as the variables. simplify your answer.)

Explanation:

Step1: Recall limit - definition

The definition of $\lim_{x
ightarrow a}f(x) = L$ is that for every $\epsilon>0$, there exists a $\delta > 0$ such that if $0<|x - a|<\delta$, then $|f(x)-L|<\epsilon$. Here $a = 2$ and $L = 6$.

Step2: Solve for $\delta$ when $\epsilon=2$

We want $|f(x)-6|<2$ whenever $0<|x - 2|<\delta$. From the graph or the limit - definition, assume $f(x)$ is a linear - like function near $x = 2$. Let's work algebraically. If $|f(x)-6|<2$, then $4

Step3: Find $\delta$ from the graph

By looking at the graph of $y = f(x)$, when $y$ is in the interval $(4,8)$, the corresponding $x$ - values satisfy $0<|x - 2|<1$. So when $\epsilon = 2$, $\delta=1$.

Step4: Solve for $\delta$ when $\epsilon = 1$

We want $|f(x)-6|<1$, i.e., $5

Step5: Make a conjecture

Based on the above results, we conjecture that $\delta=\frac{\epsilon}{2}$ for any $\epsilon>0$.

Answer:

(a) When $\epsilon = 2$, $\delta = 1$.
(b) When $\epsilon = 1$, $\delta=\frac{1}{2}$.
(c) Conjecture: $\delta=\frac{\epsilon}{2}$ for any $\epsilon>0$.