Continuity Check at x = 2

For the function $f(x)$ to be continuous at $x = 2$, the following condition must hold:

$ \lim_{x \to 2} f(x) = f(2) $

From the function definition, we know that $f(2) = k$.

Now, we need to evaluate the limit:

$ L = \lim_{x \to 2} (x-1)^\frac{1}{2-x} $

Evaluating the Limit

As $x \to 2$, the expression $(x-1)^\frac{1}{2-x}$ approaches $(2-1)^\frac{1}{2-2}$, which is $1^\frac{1}{0}$. This is an indeterminate form of the type $1^\infty$.

To resolve this, let $y = (x-1)^\frac{1}{2-x}$. Taking the natural logarithm on both sides:

$ \ln y = \ln \left( (x-1)^\frac{1}{2-x} \right) $

$ \ln y = \frac{1}{2-x} \ln(x-1) = \frac{\ln(x-1)}{2-x} $

Now, we find the limit of $\ln y$ as $x \to 2$:

$ \lim_{x \to 2} \ln y = \lim_{x \to 2} \frac{\ln(x-1)}{2-x} $

As $x \to 2$, the numerator $\ln(x-1) \to \ln(1) = 0$, and the denominator $2-x \to 0$. This is an indeterminate form of the type $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

Applying L'Hôpital's Rule

Differentiate the numerator and the denominator with respect to $x$:

• Derivative of numerator: $\frac{d}{dx}(\ln(x-1)) = \frac{1}{x-1}$
• Derivative of denominator: $\frac{d}{dx}(2-x) = -1$

Now, apply the rule:

$ \lim_{x \to 2} \ln y = \lim_{x \to 2} \frac{\frac{1}{x-1}}{-1} $

$ = \lim_{x \to 2} \frac{-1}{x-1} $

Substitute $x = 2$:

$ = \frac{-1}{2-1} = \frac{-1}{1} = -1 $

So, we found that $\lim_{x \to 2} \ln y = -1$.

Determining the Value of k

Since $\ln L = -1$, we can find $L$ by exponentiating both sides:

$ L = e^{-1} $

For continuity, $f(2) = \lim_{x \to 2} f(x)$, which means $k = L$.

Therefore, $k = e^{-1}$.