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: Take logarithms
Let $L = \lim f(x)$. Then $\ln L = \lim \dfrac{3}{2x-4}\ln\dfrac{5x-8}{8-3x}$.

Step 2: Small step
Put $x = 2 + h$. The base is $\dfrac{2 + 5h}{2 - 3h} = \dfrac{1 + 2.5h}{1 - 1.5h}$, so $\ln(\text{base}) \approx 2.5h + 1.5h = 4h$.

Step 3: Limit
The exponent is $\dfrac{3}{2h}$, so $\ln L = \dfrac{3}{2h}\cdot 4h = 6$.

Step 4: Result
$L = e^6$, so $f(2) = e^6$.

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