Question

If \(\underset{x\rightarrow 0}{lim}\frac{45^x-9^x-5^x+1}{(k^x-1)(3^x-1)} = 2\), then the value of \(k\) is ...

• A. \(45\)
• B. \(9\)
• C. \(5\)
• D. \(3\)

Hint:

Factorise the numerator as (9^x-1)(5^x-1) and use the limit of (a^x-1)/x.

Answer 1

The Correct Option is C

Step 1: Factorise
\(45^x - 9^x - 5^x + 1 = 9^x\cdot5^x - 9^x - 5^x + 1 = (9^x-1)(5^x-1)\).

Step 2: Rewrite the limit
\[ \lim_{x\to0}\frac{(9^x-1)(5^x-1)}{(k^x-1)(3^x-1)} \]
Divide the numerator and the denominator by \(x^2\).

Step 3: Use the standard limit
\(\lim_{x\to0}\frac{a^x-1}{x} = \ln a\). The limit becomes \(\frac{\ln9\cdot\ln5}{\ln k\cdot\ln3}\).

Step 4: Solve for k
\(\ln 9 = 2\ln3\), so the value is \(\frac{2\ln5}{\ln k}\). Setting this equal to \(2\) gives \(\ln k = \ln5\), so \(k=5\).

Step 5: Check
If \(k=3\) the value would be \(2\ln5/\ln3\), not 2. So option (C) is right.

Final Answer:
The value of k is 5. \[ \boxed{\text{(C)}\ 5} \]