Question

If \( f(x) = \begin{cases \dfrac{81^x - 9^x}{k^x - 1}, & x \neq 0
2, & x = 0 \end{cases} \) is continuous at \( x = 0 \), then the value of \( k \) is}

• A. \( 3 \)
• B. \( 9 \)
• C. \( 2 \)
• D. \( 4 \)

Hint:

For limits involving \( a^x-1 \), use the approximation \( a^x - 1 \approx x\ln a \) as \( x \to 0 \).

Answer 1

The Correct Option is A

Step 1: Apply continuity condition at \( x = 0 \).
For continuity, \[ \lim_{x\to 0} f(x) = f(0) = 2 \] Step 2: Evaluate the limit.
\[ \lim_{x\to 0} \frac{81^x - 9^x}{k^x - 1} \] Using \( a^x - 1 \approx x\ln a \) as \( x \to 0 \), \[ \lim_{x\to 0} \frac{x(\ln 81 - \ln 9)}{x\ln k} = \frac{\ln(9)}{\ln k} \] Step 3: Use the continuity value.
\[ \frac{\ln 9}{\ln k} = 2 \Rightarrow \ln k = \frac{\ln 9}{2} = \ln 3 \] Step 4: Solve for \( k \).
\[ k = 3 \] Step 5: Conclusion.
The value of \( k \) is \( 3 \).