Question

If the function \(f(x) = (\frac{5x-8}{8-3x})^{\frac{3}{2x-4}}\), for \(x\neq 2\) is continuous at \(x = 2\), then the value of \(f(2)\) is...

• A. \(e^{12}\)
• B. \(e^6\)
• C. \(e^3\)
• D. \(e^{\frac{3}{2}}\)

Hint:

This is a 1 to the power infinity form, so use e raised to the limit of (base - 1) times the exponent.

Answer 1

The Correct Option is B

Step 1: Understand the concept
For continuity at \(x = 2\) we need \(f(2) = \lim_{x\to2} f(x)\). At \(x = 2\) the base is \(\frac{5\cdot2 - 8}{8 - 6} = \frac{2}{2} = 1\) and the exponent \(\frac{3}{2x-4}\) tends to infinity. This is the form \(1^\infty\).

Step 2: Use the standard result
For \(1^\infty\): \(\lim \left[u(x)\right]^{v(x)} = e^{\lim v(x)(u(x) - 1)}\).

Step 3: Compute the exponent limit
\(u - 1 = \dfrac{5x - 8 - 8 + 3x}{8 - 3x} = \dfrac{8(x-2)}{8 - 3x}\). Multiply by \(\dfrac{3}{2(x-2)}\):
\[ \frac{8(x-2)}{8-3x}\cdot\frac{3}{2(x-2)} = \frac{12}{8 - 3x} \to \frac{12}{2} = 6 \]

Step 4: Result
\(f(2) = e^6\), option (B). The value \(e^{12}\) would come from forgetting the factor \(\frac{1}{2}\) in \(2x - 4 = 2(x-2)\).

Final Answer:
f(2) equals e^6. This is option (B). \[ \boxed{\text{(B) }e^6} \]