To solve this problem and determine a value of δ that satisfies the given condition:

Given:

1. $\lim_{x \to 2} f(x) = 2$.
2. The condition: If $0 < |x - 2| < \delta$, then $|f(x) - 2| < 1$.

We need to find a δ such that this condition holds true.

Steps:

1. The graph shows $f(x)$ as increasing and passing through $(2, 2)$, and the function is continuous.
2. The inequality $|f(x) - 2| < 1$ means that $f(x)$ must be between $1$ and $3$: $1 < f(x) < 3.$
3. We now examine the graph around $x = 2$ to determine where $f(x)$ remains in this range.

From the graph:

• When $x$ is close to $2$, $f(x)$ remains between $1$ and $3$ approximately within the interval $1 \leq x \leq 3$.

Thus, the distance $|x - 2|$ must satisfy: $0 < |x - 2| < 1.$

Solution:

$\delta \leq 1.$

Final Answer:

$\delta \leq 1.$

Would you like a detailed explanation of the epsilon-delta concept used here? Let me know!

Related Questions:

1. What is the definition of a limit using the epsilon-delta approach?
2. How does the graph of a function affect the choice of delta ($\delta$)?
3. Can the value of $\delta$ vary for different values of epsilon ($\epsilon$)?
4. What happens if a function is discontinuous at $x = 2$?
5. How can we generalize this approach for a more complex function?

Tip:

When working with limits and the epsilon-delta definition, always look closely at the behavior of the function near the limit point!