Question

$$\lim_{x \to 2} (x - 1)^{\frac{1}{3x - 6}} =$$

• A. $e^2$
• B. $e^3$
• C. $e^{\frac{1}{3}}$
• D. $e^{\frac{1}{2}}$

Hint:

Whenever a limit has the form $\lim_{t \to 0} (1 + t)^{\frac{k}{t}}$, its value is simply $e^k$. By rewriting our limit with a substitution $t = x - 2$, the expression becomes $\lim_{t \to 0} (1 + t)^{\frac{1}{3t}}$, which directly gives $e^{\frac{1}{3}}$ in one step!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are asked to evaluate the limit of an exponential functional structure $\lim_{x \to 2} (x - 1)^{\frac{1}{3x - 6}}$. Direct substitution of $x = 2$ gives the indeterminate form $1^\infty$.

Step 2: Key Formula or Approach:
For any limit yielding a standard $1^\infty$ indeterminate form where $\lim_{x \to c} f(x) = 1$ and $\lim_{x \to c} g(x) = \infty$, the standard evaluation shortcut rule states: $$\lim_{x \to c} [f(x)]^{g(x)} = e^{\lim_{x \to c} [f(x) - 1] \cdot g(x)}$$ Here, $f(x) = x - 1$ and $g(x) = \frac{1}{3x - 6}$.

Step 3: Detailed Explanation:
Let's apply our standard rule directly to convert the expression to base $e$: $$L = e^{\lim_{x \to 2} [(x - 1) - 1] \cdot \left(\frac{1}{3x - 6}\right)}$$ Simplify the algebraic term in the exponent: $$\text{Exponent} = \lim_{x \to 2} (x - 2) \cdot \frac{1}{3(x - 2)}$$ Notice that the factor $(x - 2)$ is present in both the numerator and the denominator. Since $x \to 2$, $x \neq 2$, so we can cancel $(x - 2)$ completely: $$\text{Exponent} = \lim_{x \to 2} \frac{1}{3} = \frac{1}{3}$$ Substituting this constant value back into our base $e$ expression gives: $$L = e^{\frac{1}{3}}$$

Step 4: Final Answer:
The evaluated value of the limit is $e^{\frac{1}{3}}$, corresponding to option (C).