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 (type an inequality using δ as the variable. 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, we need to find the $x$-values around $x = 2$ for which the $y$-values of the function $y = f(x)$ satisfy $|y - 6|<2$, i.e., $4By looking at the graph, we find the $x$-values corresponding to $y = 4$ and $y = 8$. Let's assume the function is linear - like in the neighborhood of $x = 2$.
If we consider the points on the graph, we find that when $y = 4$, $x$ is some value $x_1$ and when $y = 8$, $x$ is some value $x_2$.
The distance from $x = 2$ to the $x$-values where $y$ is at the boundaries of the $\epsilon$-interval gives us $\delta$.
From the graph, when $\epsilon=2$, we find that $\delta = 1$. So the inequality is $\delta\leq1$.

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

We want $|f(x)-6|<1$ whenever $0 < |x - 2|<\delta$. So we need to find the $x$-values around $x = 2$ for which $5By looking at the graph, we find the $x$-values corresponding to $y = 5$ and $y = 7$.
The distance from $x = 2$ to these $x$-values gives us $\delta$. From the graph, when $\epsilon = 1$, $\delta=\frac{1}{2}$. So the inequality is $\delta\leq\frac{1}{2}$.

Step4: Make a conjecture

As $\epsilon$ gets smaller, $\delta$ also gets smaller. In general, for any $\epsilon>0$, $\delta$ is chosen such that the part of the graph of $y = f(x)$ for $x$ in the interval $(2-\delta,2 + \delta)\setminus\{2\}$ lies within the horizontal strip $y\in(6-\epsilon,6+\epsilon)$. We can conjecture that $\delta=\frac{\epsilon}{2}$ (assuming the function is linear - like in the neighborhood of $x = 2$).

Answer:

(a) $\delta\leq1$
(b) $\delta\leq\frac{1}{2}$
(c) $\delta=\frac{\epsilon}{2}$