To determine for what values of [tex]\( k \)[/tex] the function [tex]\( f \)[/tex] defined by
[tex]\[
f(x) = \begin{cases}
\frac{|x-3|}{x-3}, & \text{if } x > 3 \\
k, & \text{if } x \leq 3
\end{cases}
\][/tex]
is continuous at [tex]\( x = 3 \)[/tex], we need to ensure the limit from the left and the limit from the right at [tex]\( x = 3 \)[/tex] are equal, and also equal to the function value at [tex]\( x = 3 \)[/tex].
First, let's simplify [tex]\( \frac{|x-3|}{x-3} \)[/tex] for [tex]\( x > 3 \)[/tex]:
For [tex]\( x > 3 \)[/tex]:
[tex]\[
|x-3| = x-3 \implies \frac{|x-3|}{x-3} = \frac{x-3}{x-3} = 1.
\][/tex]
Thus, for [tex]\( x > 3 \)[/tex], [tex]\( f(x) = 1 \)[/tex].
Now, the function can be rewritten as:
[tex]\[
f(x) = \begin{cases}
1, & \text{if } x > 3 \\
k, & \text{if } x \leq 3
\end{cases}
\][/tex]
To check continuity at [tex]\( x = 3 \)[/tex], we need the following to hold:
1. [tex]\(\lim_{x \to 3^-} f(x)\)[/tex] exists.
2. [tex]\(\lim_{x \to 3^+} f(x)\)[/tex] exists.
3. [tex]\(f(3)\)[/tex] is defined.
4. [tex]\(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)\)[/tex].
Let's calculate each condition:
1. [tex]\[
\lim_{x \to 3^-} f(x) \rightarrow \lim_{x \to 3^-} k = k.
\][/tex]
2. [tex]\[
\lim_{x \to 3^+} f(x) \rightarrow \lim_{x \to 3^+} 1 = 1.
\][/tex]
3. [tex]\[
f(3) = k.
\][/tex]
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 3 \)[/tex]:
[tex]\[
\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3).
\][/tex]
Therefore,
[tex]\[
k = 1.
\][/tex]
As a result, the function [tex]\( f \)[/tex] is continuous at [tex]\( x = 3 \)[/tex] if and only if [tex]\( k = 1 \)[/tex].
Hence, the value of [tex]\( k \)[/tex] is 1, which is unfortunately not listed among the options.
For the given answer for the limit problem
[tex]\[
\lim_{x \rightarrow 0} \frac{\sin(7x) - \sin(4x)}{3x} = 1.
\][/tex]
Let's analyze the problem:
Using L'Hopital's Rule for the limit,
[tex]\[
=\lim_{x \rightarrow 0} \frac{7\cos(7x) - 4\cos(4x)}{3}.
\][/tex]
When [tex]\( x \to 0\)[/tex],
[tex]\[
=\frac{7\cos(0) - 4\cos(0)}{3} = \frac{7 - 4}{3} = 1.
\][/tex]
Thus, the correct answer is [tex]\( \boxed{1} \)[/tex].
