Let's solve each part of the question step by step:
(a) Finding [tex]\(\lim_{{x \to 2^{-}}} f(x)\)[/tex] and [tex]\(\lim_{{x \to 2^{+}}} f(x)\)[/tex]
1. Left-hand limit as [tex]\(x\)[/tex] approaches 2 ([tex]\(\lim_{{x \to 2^{-}}} f(x)\)[/tex]):
For [tex]\(x \leq 2\)[/tex], the function [tex]\(f(x)\)[/tex] is defined as [tex]\(4 - x^2\)[/tex]. To find the left-hand limit, we find the value of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 2 from the left:
[tex]\[
\lim_{{x \to 2^{-}}} f(x) = \lim_{{x \to 2^{-}}} (4 - x^2)
\][/tex]
Substituting [tex]\(x = 2\)[/tex]:
[tex]\[
\lim_{{x \to 2^{-}}} (4 - x^2) = 4 - 2^2 = 4 - 4 = 0
\][/tex]
Therefore,
[tex]\[
\lim_{{x \to 2^{-}}} f(x) = 0
\][/tex]
2. Right-hand limit as [tex]\(x\)[/tex] approaches 2 ([tex]\(\lim_{{x \to 2^{+}}} f(x)\)[/tex]):
For [tex]\(x > 2\)[/tex], the function [tex]\(f(x)\)[/tex] is defined as [tex]\(x - 1\)[/tex]. To find the right-hand limit, we find the value of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 2 from the right:
[tex]\[
\lim_{{x \to 2^{+}}} f(x) = \lim_{{x \to 2^{+}}} (x - 1)
\][/tex]
Substituting [tex]\(x = 2\)[/tex]:
[tex]\[
\lim_{{x \to 2^{+}}} (x - 1) = 2 - 1 = 1
\][/tex]
Therefore,
[tex]\[
\lim_{{x \to 2^{+}}} f(x) = 1
\][/tex]
(b) Determining if [tex]\(\lim_{{x \to 2}} f(x)\)[/tex] exists
For the limit [tex]\(\lim_{{x \to 2}} f(x)\)[/tex] to exist, the left-hand limit and the right-hand limit must be equal. From part (a), we have:
[tex]\[
\lim_{{x \to 2^{-}}} f(x) = 0
\][/tex]
[tex]\[
\lim_{{x \to 2^{+}}} f(x) = 1
\][/tex]
Since [tex]\(0 \ne 1\)[/tex], the left-hand limit and the right-hand limit are not equal. Hence,
[tex]\[
\lim_{{x \to 2}} f(x) \text{ does not exist}
\][/tex]
(c) Sketching the graph of [tex]\(f(x)\)[/tex]
To sketch the graph of the piecewise function [tex]\(f(x)\)[/tex], we need to consider each piece of the function individually:
1. For [tex]\(x \leq 2\)[/tex], the function is [tex]\(f(x) = 4 - x^2\)[/tex]:
- This is part of a downward-opening parabola with vertex at [tex]\((0, 4)\)[/tex].
- At [tex]\(x = 2\)[/tex], [tex]\(f(2) = 4 - 2^2 = 0\)[/tex].
2. For [tex]\(x > 2\)[/tex], the function is [tex]\(f(x) = x - 1\)[/tex]:
- This is a linear function with slope 1 and y-intercept [tex]\(-1\)[/tex].
- At [tex]\(x = 2\)[/tex], for the right-hand limit, the value is [tex]\(1\)[/tex] (even though it is not included in the actual function).
- As [tex]\(x\)[/tex] increases beyond [tex]\(2\)[/tex], [tex]\(f(x)\)[/tex] increases linearly.
When combining these insights into one graph, we notice:
- There is a solid point at [tex]\((2, 0)\)[/tex] on the left parabola because the function includes this point for [tex]\(x \leq 2\)[/tex].
- There is an open circle at [tex]\((2, 1)\)[/tex] because [tex]\(x = 2\)[/tex] is not included in the line segment, but the right-hand value approaches 1.
Below is a rough sketch of the graph based on our analysis:
```plaintext
y
|
4|
| / \
3| / \
| / \
2| / \
| / \
1|o-----------
| \
0|.............\..........
-1|----------------------------
-2 -1 0 1 2 3 4 5 x
```
Note:
- The parabola segment [tex]\((4 - x^2)\)[/tex] applies for [tex]\(x \leq 2\)[/tex].
- The linear segment [tex]\((x - 1)\)[/tex] starts from just right of [tex]\(x = 2\)[/tex], drawn as a dashed line because [tex]\(x = 2\)[/tex] itself is not part of this segment in the piecewise function.
