Question

If \(f(x)\) is continuous at \(x = 0\), where \(f(x) = \frac{8^x-2^x}{k^x-1}\), for \(x\neq 0\) and \(f(0) = 2\), then the value of \(k\) is ...

• A. \(0\)
• B. \(4\)
• C. \(-2\)
• D. \(2\)

Hint:

Continuity needs the limit at 0 to equal 2; factor \(2^x\) from the numerator and use \(\lim\frac{a^x-1}{x}=\ln a\).

Answer 1

The Correct Option is D

Step 1: Plan:
Apply L'Hopital's rule, because the form is $\frac00$.

Step 2: Steps:
Differentiate top and bottom: $\frac{8^x\ln8 - 2^x\ln2}{k^x\ln k}$. At $x = 0$ this is $\frac{\ln8-\ln2}{\ln k} = \frac{\ln4}{\ln k}$.
Set it equal to $2$: $\ln k = \ln 2$, so $k = 2$.

Final Answer:
The value of $k$ is $2$, option (D). \[ \boxed{2} \]