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Step 1: Understand the Problem
We are given a function
$f(x) = \begin{cases}
\frac{\sqrt{2}\,\cos x - 1}{\cot x - 1}, & x \neq \frac{\pi}{4} \\
k, & x = \frac{\pi}{4}
\end{cases}$
defined on the interval $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$. We need to determine the value of $k$ if $f(x)$ is continuous at $x = \frac{\pi}{4}$.
Step 2: Continuity Condition
For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, the limit of $f(x)$ as $x$ approaches $\frac{\pi}{4}$ must be equal to the value of the function at $\frac{\pi}{4}$. In other words,
$$
\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\,\cos x - 1}{\cot x - 1} \;=\; f\!\bigl(\tfrac{\pi}{4}\bigr) \;=\; k.
$$
Step 3: Check the Form of the Limit
Before applying any method, we check what happens if we directly substitute $x = \frac{\pi}{4}$ into the expression
$\frac{\sqrt{2}\,\cos x - 1}{\cot x - 1}$.
At $x=\frac{\pi}{4}$:
$\cos \tfrac{\pi}{4} = \frac{1}{\sqrt{2}}$, so $\sqrt{2}\,\cos \tfrac{\pi}{4} - 1 = \sqrt{2}\times\frac{1}{\sqrt{2}} - 1 = 1 - 1 = 0.$
$\cot \tfrac{\pi}{4} = 1$, so $\cot \tfrac{\pi}{4} - 1 = 1 - 1 = 0.$
Thus, the limit expression is of the indeterminate form $\tfrac{0}{0}$. This means we can apply L'Hôpital's Rule.
Step 4: Apply L'Hôpital's Rule
L'Hôpital's Rule says that for a limit of the form $\frac{0}{0}$, we can take derivatives of the numerator and denominator (with respect to $x$) and then re-evaluate the limit. So we compute:
Numerator: $\sqrt{2}\,\cos x - 1 \;\to\; \frac{d}{dx}(\sqrt{2}\,\cos x - 1) = -\sqrt{2}\,\sin x.$
Denominator: $\cot x - 1 \;\to\; \frac{d}{dx}(\cot x - 1) = -\csc^2 x.$
Therefore,
$$
\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\,\cos x - 1}{\cot x - 1}
=
\lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2}\,\sin x}{-\csc^2 x}
=
\lim_{x \to \frac{\pi}{4}} \sqrt{2}\,\sin x\,\sin^2 x
=
\lim_{x \to \frac{\pi}{4}} \sqrt{2}\,\sin^3 x.
$$
Step 5: Evaluate the Simplified Limit
Now we substitute $x = \frac{\pi}{4}$ into $\sqrt{2}\,\sin^3 x$:
$$
\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}},
\quad
\text{so}
\quad
\sin^3 \frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{1}{2\sqrt{2}}.
$$
Thus,
$$
\sqrt{2} \,\sin^3 \frac{\pi}{4}
=
\sqrt{2} \times \frac{1}{2\sqrt{2}}
=
\frac{1}{2}.
$$
Step 6: Conclude the Value of k
This limit must equal $k$ for continuity, so we have:
$$
k = \frac{1}{2}.
$$
