Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x = a$ if and only if the following three conditions are met:

  1. $f(a)$ is defined.
  2. $\mathop {\lim }\limits_{x \to a} f(x)$ exists.
  3. $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$.

• Limit of the form $1^\infty$: To evaluate a limit of the form $1^\infty$, say $\mathop {\lim }\limits_{x \to a} {[g(x)]}^{h(x)}$, where $\mathop {\lim }\limits_{x \to a} g(x) = 1$ and $\mathop {\lim }\limits_{x \to a} h(x) = \infty$, we can use the formula: $\mathop {\lim }\limits_{x \to a} {[g(x)]}^{h(x)} = e^{\mathop {\lim }\limits_{x \to a} {h(x)[g(x) - 1]}}$

Step-by-Step Solution

Step 1: Understand the condition for continuity. The problem states that the function $f(x)$ is continuous at $x = 2$. For $f(x)$ to be continuous at $x = 2$, the limit of the function as $x$ approaches 2 must be equal to the value of the function at $x = 2$. Mathematically, this is expressed as: $\mathop {\lim }\limits_{x \to 2} f(x) = f(2)$

Step 2: Identify the values of $f(2)$ and the limit expression. From the definition of the function, we are given that $f(2) = k$. For the limit, as $x$ approaches 2, we use the first part of the function definition, where $x > 1$ and $x \ne 2$: $f(x) = (x - 1)^{\frac{1}{2 - x}}$ Therefore, the continuity condition becomes: $\mathop {\lim }\limits_{x \to 2} (x - 1)^{\frac{1}{2 - x}} = k$

Step 3: Evaluate the limit. Let's evaluate the limit: $L = \mathop {\lim }\limits_{x \to 2} (x - 1)^{\frac{1}{2 - x}}$. As $x \to 2$, the base $(x - 1) \to (2 - 1) = 1$. As $x \to 2$, the exponent $\frac{1}{2 - x}$ approaches $\frac{1}{0}$. If $x \to 2^+$, the denominator $2-x \to 0^-$, so $\frac{1}{2-x} \to -\infty$. If $x \to 2^-$, the denominator $2-x \to 0^+$, so $\frac{1}{2-x} \to +\infty$. Thus, the limit is of the indeterminate form $1^\infty$.

Step 4: Apply the formula for the $1^\infty$ indeterminate form. We use the formula: $\mathop {\lim }\limits_{x \to a} {[g(x)]}^{h(x)} = e^{\mathop {\lim }\limits_{x \to a} {h(x)[g(x) - 1]}}$. Here, $g(x) = x - 1$ and $h(x) = \frac{1}{2 - x}$, and $a = 2$. So, the limit $L$ is given by: $L = e^{\mathop {\lim }\limits_{x \to 2} {\frac{1}{2 - x}}[(x - 1) - 1]}$ $L = e^{\mathop {\lim }\limits_{x \to 2} {\frac{1}{2 - x}}[x - 2]}$ $L = e^{\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}}}$

Step 5: Simplify the exponent and find its limit. The expression in the exponent is $\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}}$. We can rewrite the numerator as $-(2 - x)$. $\mathop {\lim }\limits_{x \to 2} {\frac{-(2 - x)}{2 - x}}$ For $x \ne 2$, we can cancel out the $(2 - x)$ term: $\mathop {\lim }\limits_{x \to 2} {-1} = -1$

Step 6: Determine the value of k. Now substitute the limit of the exponent back into the expression for $L$: $L = e^{-1}$ Since the function is continuous at $x = 2$, we have $L = f(2) = k$. Therefore, $k = e^{-1}$.

Step 7: Recheck the calculation and the problem statement. Let's review the steps. The limit form is $1^\infty$. The formula applied is correct. The simplification of the exponent is: $\mathop {\lim }\limits_{x \to 2} {\frac{1}{2 - x}}[(x - 1) - 1] = \mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}} = \mathop {\lim }\limits_{x \to 2} {\frac{-(2 - x)}{2 - x}} = \mathop {\lim }\limits_{x \to 2} {-1} = -1$. So, $k = e^{-1}$.

However, the provided correct answer is (A) 1. This suggests a potential misinterpretation or a common error in the original solution provided in the problem description. Let's re-examine the problem and the standard methods.

Let's re-evaluate the limit using a logarithmic approach, which is often clearer. Let $y = (x - 1)^{\frac{1}{2 - x}}$. Taking the natural logarithm of both sides: $\ln y = \ln \left( (x - 1)^{\frac{1}{2 - x}} \right)$ $\ln y = \frac{1}{2 - x} \ln(x - 1)$ Now, we find the limit of $\ln y$ as $x \to 2$: $\mathop {\lim }\limits_{x \to 2} \ln y = \mathop {\lim }\limits_{x \to 2} \frac{\ln(x - 1)}{2 - x}$ As $x \to 2$, the numerator $\ln(x - 1) \to \ln(2 - 1) = \ln(1) = 0$. As $x \to 2$, the denominator $2 - x \to 0$. This is of the indeterminate form $\frac{0}{0}$, so we can use L'Hôpital's Rule.

Applying L'Hôpital's Rule: $\mathop {\lim }\limits_{x \to 2} \frac{\frac{d}{dx}(\ln(x - 1))}{\frac{d}{dx}(2 - x)} = \mathop {\lim }\limits_{x \to 2} \frac{\frac{1}{x - 1}}{-1}$ $= \mathop {\lim }\limits_{x \to 2} \frac{1}{-1(x - 1)}$ $= \frac{1}{-1(2 - 1)}$ $= \frac{1}{-1(1)}$ $= -1$ So, $\mathop {\lim }\limits_{x \to 2} \ln y = -1$. Since $\ln y \to -1$, then $y \to e^{-1}$. Thus, $k = e^{-1}$.

There seems to be a discrepancy with the provided correct answer. Let's consider if there was a typo in the problem or the options.

Let's re-examine the original solution provided: "$\therefore$ ${{e^l}}$ = k Where $l$ = $\mathop {\lim }\limits_{x \to 2}$(x $-$ $l$ $-$ 1) $\times$ ${1 \over {2 - x}}$ = $\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$ $\Rightarrow$ k = e $-$1"

The original solution has a typo in the calculation of $l$: "($(x - 1) - 1$)" which is correct, but then it skips the step and directly writes "$\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$". This part of the calculation is correct. The conclusion "$k = e^{-1}$" is also correct based on this calculation.

However, the provided correct answer is (A) 1. This implies that $k=1$. For $k=1$, we would need $e^{-1} = 1$, which is false, or if the limit of the exponent was 0, then $e^0 = 1$. Let's check if the exponent limit could be 0.

If the function was something like $(x-1)^{\frac{1}{x-2}}$ as $x \to 2$, the exponent would be $\frac{1}{0}$ which is $\pm \infty$. If the function was $(x-1)^{\frac{1}{(x-2)^2}}$ as $x \to 2$, the exponent would be $\frac{1}{0^+}$ which is $+\infty$.

Let's assume there might be a typo in the question and consider a scenario that leads to $k=1$. If the function was $f(x) = (x-1)^{\frac{1}{x-2}}$ for $x>1, x \ne 2$. Then $\mathop {\lim }\limits_{x \to 2} (x-1)^{\frac{1}{x-2}}$. This is $1^\infty$ form. The exponent term is $\mathop {\lim }\limits_{x \to 2} \frac{1}{x-2} ((x-1)-1) = \mathop {\lim }\limits_{x \to 2} \frac{x-2}{x-2} = 1$. In this case, $k = e^1 = e$. This is option (B).

Let's consider another possibility. What if the base was slightly different? Suppose the function was $f(x) = (2-x)^{\frac{1}{2-x}}$ as $x \to 2$. This is not defined for $x>1$ and $x \ne 2$.

Let's strictly follow the given function and aim for the provided correct answer (A) $k=1$. For $k=1$, we need $\mathop {\lim }\limits_{x \to 2} (x - 1)^{\frac{1}{2 - x}} = 1$. This means $e^{\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}}} = 1$. For $e^A = 1$, we need $A = 0$. So, we would need $\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}} = 0$. But we calculated this limit to be $-1$.

This indicates a strong possibility that the provided "Correct Answer: A" is incorrect for the given question. Based on standard limit evaluation techniques, the limit is $e^{-1}$.

Let's re-examine the original solution's calculation for the exponent: $l = \mathop {\lim }\limits_{x \to 2} (x - 1 - 1) \times \frac{1}{2 - x}$ $l = \mathop {\lim }\limits_{x \to 2} (x - 2) \times \frac{1}{2 - x}$ $l = \mathop {\lim }\limits_{x \to 2} \frac{x - 2}{2 - x}$ $l = \mathop {\lim }\limits_{x \to 2} \frac{-(2 - x)}{2 - x}$ $l = -1$ Then $k = e^l = e^{-1}$.

If the question intended for the answer to be 1, then the exponent limit must be 0. Perhaps the function was meant to be $f(x) = (x-1)^{\frac{x-2}{2-x}}$? Then the exponent would be $\mathop {\lim }\limits_{x \to 2} \frac{x-2}{2-x} \times \frac{x-2}{2-x} = \mathop {\lim }\limits_{x \to 2} (-1) \times (-1) = 1$. So $k = e^1 = e$.

What if the function was $f(x) = (x-1)^{\frac{-(x-2)}{2-x}}$? Then the exponent would be $\mathop {\lim }\limits_{x \to 2} \frac{-(x-2)}{2-x} \times \frac{x-2}{2-x} = \mathop {\lim }\limits_{x \to 2} (-1) \times (-1) = 1$. So $k=e$.

Let's consider a scenario where the exponent becomes zero. If the function was $f(x) = (x-1)^{g(x)}$ and $\mathop {\lim }\limits_{x \to 2} g(x) = 0$. For example, if $g(x) = (x-2) \times \frac{1}{2-x} \times (something \to 0)$.

Given the strong contradiction between the derived answer and the provided correct answer, it is highly probable that the provided correct answer is incorrect. However, as per the instructions, I must work towards the provided answer if possible.

Let's assume the question implies some non-standard interpretation or a very subtle point. Could the limit of the exponent be 0 in some way? The expression is $\mathop {\lim }\limits_{x \to 2} \frac{x-2}{2-x}$. This is definitively -1.

Let's consider a common mistake that might lead to 1. Perhaps the formula for $1^\infty$ was misapplied, or the base was assumed to be exactly 1. If $\mathop {\lim }\limits_{x \to 2} f(x) = 1$. This would mean $k=1$.

Let's assume that the question is flawed and the correct answer should be $e^{-1}$. If we are forced to choose from the options and the correct answer is indeed (A) 1, then there must be an error in our limit calculation or the problem statement.

Let's review the problem statement very carefully. f(x) = \left\{ {\matrix{ {{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr } } \right.

The calculation of the limit $\mathop {\lim }\limits_{x \to 2} (x - 1)^{\frac{1}{2 - x}}$ as $e^{-1}$ is robust.

Given the constraint to reach the provided correct answer, and the impossibility of doing so with standard calculus methods for the given function, it is impossible to provide a step-by-step derivation that logically leads to $k=1$.

However, if we were to assume that the limit should evaluate to 1, the only way for $k=1$ to be the answer is if $\mathop {\lim }\limits_{x \to 2} f(x) = 1$. This would imply $e^{\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}}} = 1$. This requires $\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}} = 0$. But $\mathop {\lim }\limits_{x \to 2} {\frac{x - 2}{2 - x}} = -1$.

There seems to be an error in the provided correct answer. The mathematically derived answer is $e^{-1}$.

If, hypothetically, the question was designed such that the exponent's limit was 0, then $k$ would be $e^0 = 1$. For the exponent $\frac{1}{2-x} (x-1-1)$ to have a limit of 0, the term $\frac{x-2}{2-x}$ would need to have a limit of 0. This is not possible.

Let's consider if there's a misunderstanding of the question or a very common trick. The question is standard for limits and continuity.

Given the situation, and the strict instruction to reach the provided correct answer, I cannot provide a valid step-by-step derivation for $k=1$. The correct derivation leads to $k=e^{-1}$.

If I were forced to reverse-engineer a scenario where $k=1$: This would imply $\mathop {\lim }\limits_{x \to 2} f(x) = 1$. So, $\mathop {\lim }\limits_{x \to 2} (x - 1)^{\frac{1}{2 - x}} = 1$. This means $e^{\mathop {\lim }\limits_{x \to 2} \frac{1}{2-x}((x-1)-1)} = 1$. $e^{\mathop {\lim }\limits_{x \to 2} \frac{x-2}{2-x}} = 1$. $e^{-1} = 1$. This is false.

It is highly likely that the provided correct answer (A) 1 is incorrect for the given problem statement. The correct answer based on standard calculus is $e^{-1}$.

However, to fulfill the request of reaching the given correct answer, I must assume there's an error in my understanding or the problem statement is subtly different. Since I cannot find any such error that leads to 1, I cannot proceed.

Let's assume, for the sake of argument, that the question had a typo and the function was intended to produce a limit of 1. The most straightforward way to get a limit of 1 from a $1^\infty$ form is if the exponent's limit is 0.

If the exponent was, for example, $\frac{x-2}{2-x} \times (x-2)$, then as $x \to 2$, this would be $(-1) \times 0 = 0$. Then $k = e^0 = 1$. However, this is not the function given.

Given the instructions, and the impossibility of reaching the provided correct answer through valid mathematical steps from the given problem, I cannot complete this task as requested. The derived answer is $e^{-1}$.

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is in the $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate form before applying L'Hôpital's Rule.
• Algebraic errors in simplification: Carefully simplify expressions, especially when dealing with exponents and fractions, to avoid mistakes.
• Misremembering the $1^\infty$ formula: The formula for limits of the form $1^\infty$ is $e^{\mathop {\lim }\limits_{x \to a} {h(x)[g(x) - 1]}}$, not just the limit of the exponent itself.

Summary

To find the value of $k$ for which the function $f(x)$ is continuous at $x = 2$, we must equate the limit of $f(x)$ as $x$ approaches 2 with $f(2)$. The limit is of the indeterminate form $1^\infty$, which is evaluated using the formula $e^{\mathop {\lim }\limits_{x \to a} {h(x)[g(x) - 1]}}$. After applying this formula and simplifying, the limit was found to be $e^{-1}$. Therefore, $k = e^{-1}$.

Final Answer

Based on the rigorous application of calculus principles, the value of $k$ for which $f(x)$ is continuous at $x=2$ is $e^{-1}$. There appears to be an inconsistency with the provided correct answer options and the derived result.

If we were forced to select an answer and suspect a typo in the question or options, the most common intended answer for this type of problem structure, if it were to yield 1, would require the exponent's limit to be 0. However, for the given function, the limit of the exponent is -1.

The final answer is $\boxed{e^{-1}}$.