If the function $$f$$ defined on $$\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$$ by $$f(x) = \begin{cases} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$$ is continuous, then $$k$$ is equal to:

We have a real-valued function defined on the open interval $$\left(\dfrac{\pi}{6},\dfrac{\pi}{3}\right)$$ by

$$ f(x)= \begin{cases} \dfrac{\sqrt{2}\cos x-1}{\cot x-1}, & x\neq\dfrac{\pi}{4}\\[4pt] k, & x=\dfrac{\pi}{4} \end{cases} $$

For continuity at the point $$x=\dfrac{\pi}{4}$$ we require the limit of the first expression as $$x\to\dfrac{\pi}{4}$$ to exist and to be equal to $$k$$, that is

$$ k=\lim_{x\to\pi/4}\dfrac{\sqrt{2}\cos x-1}{\cot x-1}. $$

We now evaluate this limit step by step. First we rewrite the denominator in terms of sine and cosine only. Remember the definition $$\cot x=\dfrac{\cos x}{\sin x}$$. Substituting this, we obtain

$$ \cot x-1=\dfrac{\cos x}{\sin x}-1=\dfrac{\cos x-\sin x}{\sin x}. $$

Putting this back into the fraction gives

$$ f(x)=\dfrac{\sqrt{2}\cos x-1}{\dfrac{\cos x-\sin x}{\sin x}} =\dfrac{\bigl(\sqrt{2}\cos x-1\bigr)\sin x}{\cos x-\sin x}. $$

So the required limit becomes

$$ \lim_{x\to\pi/4}\dfrac{\bigl(\sqrt{2}\cos x-1\bigr)\sin x}{\cos x-\sin x}. $$

Notice that at $$x=\dfrac{\pi}{4}$$ both the numerator and the denominator separately go to zero, because

$$ \sqrt{2}\cos\!\left(\dfrac{\pi}{4}\right)-1=\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}-1=1-1=0 $$

and

$$ \cos\!\left(\dfrac{\pi}{4}\right)-\sin\!\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}=0. $$

Hence the fraction is of the indeterminate type $$\dfrac{0}{0}$$, so we apply the L’Hospital rule. The rule states that if the limits of numerator and denominator give $$0/0$$, then

$$ \lim_{x\to a}\dfrac{u(x)}{v(x)} =\lim_{x\to a}\dfrac{u'(x)}{v'(x)}, $$

provided that the latter limit exists. We therefore differentiate the numerator and the denominator separately.

Let

$$ U(x)=\bigl(\sqrt{2}\cos x-1\bigr)\sin x,\qquad V(x)=\cos x-\sin x. $$

Differentiating $$U(x)$$ with respect to $$x$$, we use the product rule $$\dfrac{d}{dx}\bigl[p(x)q(x)\bigr]=p'(x)q(x)+p(x)q'(x)$$. Writing $$p(x)=\sqrt{2}\cos x-1$$ and $$q(x)=\sin x$$, we get

$$ U'(x)=\bigl[-\sqrt{2}\sin x\bigr]\sin x+\bigl(\sqrt{2}\cos x-1\bigr)\cos x =-\sqrt{2}\sin^{2}x+\bigl(\sqrt{2}\cos x-1\bigr)\cos x. $$

Next we differentiate $$V(x)$$. Using $$\dfrac{d}{dx}\cos x=-\sin x$$ and $$\dfrac{d}{dx}\sin x=\cos x$$, we have

$$ V'(x)=-\sin x-\cos x. $$

Now we evaluate these derivatives at $$x=\dfrac{\pi}{4}$$. Recall that

$$ \sin\!\left(\dfrac{\pi}{4}\right)=\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}. $$

Therefore, for the numerator derivative:

$$ U'\!\left(\dfrac{\pi}{4}\right)= -\sqrt{2}\left(\dfrac{\sqrt{2}}{2}\right)^2 +\Bigl(\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}-1\Bigr)\dfrac{\sqrt{2}}{2} =-\sqrt{2}\cdot\dfrac12+\bigl(1-1\bigr)\dfrac{\sqrt{2}}{2} =-\dfrac{\sqrt{2}}{2}. $$

(The second term vanishes because $$\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}-1=1-1=0$$.)

For the denominator derivative:

$$ V'\!\left(\dfrac{\pi}{4}\right)= -\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2} =-\sqrt{2}. $$

Hence, by L’Hospital rule, the required limit is

$$ \lim_{x\to\pi/4}\dfrac{U(x)}{V(x)} =\dfrac{U'\!\left(\dfrac{\pi}{4}\right)}{V'\!\left(\dfrac{\pi}{4}\right)} =\dfrac{-\dfrac{\sqrt{2}}{2}}{-\sqrt{2}} =\dfrac{1}{2}. $$

Thus we must choose

$$ k=\dfrac12. $$

Hence, the correct answer is Option A.