the function f in the given figure satisfies lim f(x)=6 as x→2. for each value of ε, find all values of δ>0…

the function f in the given figure satisfies lim f(x)=6 as x→2. 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. (type an inequality using δ as the variable. simplify your answer.)

the function f in the given figure satisfies lim f(x)=6 as x→2. 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. (type an inequality using δ as the variable. simplify your answer.)

Answer

Explanation:

Step1: Recall the definition of the limit

The limit $\lim_{x\rightarrow a}f(x) = L$ means 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 $0<|x - 2|<\delta$. Since we don't know the function $f(x)$, but from the limit - definition, we assume a linear - like behavior near $x = 2$. In the case of a well - behaved function, we can often find $\delta$ by working backward from the inequality $|f(x)-6|<\epsilon$. For a simple case (assuming the function is "nice"), if we assume a linear relationship, we set $\delta=\epsilon$. So when $\epsilon = 2$, $\delta=2$.

Step3: Solve for $\delta$ when $\epsilon = 1$

Similarly, when $\epsilon=1$, following the limit definition and assuming a well - behaved function near $x = 2$, we set $\delta = 1$.

Step4: Make a conjecture

For any $\epsilon>0$, we conjecture that $\delta=\epsilon$ will satisfy the condition that $|f(x)-6|<\epsilon$ whenever $0<|x - 2|<\delta$ for a well - behaved function $f(x)$ near $x = 2$.

Answer:

(a) $\delta = 2$ (b) $\delta = 1$ (c) Conjecture: $\delta=\epsilon$