To determine which statements about the function [tex]\( f(x) \)[/tex] are true, let's analyze each one by one.
Given the function:
[tex]\[
f(x)=\left\{
\begin{array}{ll}
\frac{x^2-4}{x-2} & \text{if } x \neq 2 \\
1 & \text{if } x=2
\end{array}
\right.
\][/tex]
To simplify the expression for [tex]\( f(x) \)[/tex] when [tex]\( x \neq 2 \)[/tex]:
[tex]\[
f(x) = \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad \text{for } x \neq 2
\][/tex]
1. Limit at [tex]\( x = 2 \)[/tex]
To find the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 2 from both sides:
[tex]\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2)
= 4
\][/tex]
This implies the limit exists and is 4. However, the given [tex]\( f(x) \)[/tex] at [tex]\( x = 2 \)[/tex] is 1, not 4. Thus, the first statement is false.
[tex]\[
\text{Statement I: } f \text{ has a limit at } x = 2 \text{ is False.}
\][/tex]
2. Continuity at [tex]\( x = 2 \)[/tex]
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 2 \)[/tex]:
[tex]\[
\lim_{x \to 2} f(x) \text{ must equal } f(2)
\][/tex]
From our earlier evaluation, we know:
[tex]\[
\lim_{x \to 2} f(x) = 4 \quad \text{and} \quad f(2) = 1
\][/tex]
Since [tex]\( 4 \neq 1 \)[/tex], [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 2 \)[/tex].
[tex]\[
\text{Statement II: } f \text{ is continuous at } x = 2 \text{ is False.}
\][/tex]
3. Differentiability at [tex]\( x = 2 \)[/tex]
For [tex]\( f(x) \)[/tex] to be differentiable at [tex]\( x = 2 \)[/tex], it must be continuous there, and the derivative must exist at that point. Since we already established that [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 2 \)[/tex], [tex]\( f(x) \)[/tex] cannot be differentiable at [tex]\( x = 2 \)[/tex]. However, since [tex]\( f(x) \)[/tex] when [tex]\( x \neq 2 \)[/tex] simplifies to [tex]\( x + 2 \)[/tex], its derivative would be 1.
In [tex]\( f \)[/tex] being differentiable, there is no discontinuity in the evaluated derivative.
[tex]\[
\text{Statement III: } f \text{ is differentiable at } x = 2 \text{ is True.}
\][/tex]
To conclude, the results are:
- Statement I: False,
- Statement II: False,
- Statement III: True
The correct answer is:
[tex]\[
(C) \text{III only}
\][/tex]
