Question

Let \(f: R \to (0, \infty)\) be a twice differentiable function such that \(f(3) = 18\), \(f'(3)=0\) and \(f''(3) = 4\). Then \(\lim_{x \to 1} \log_e \left[ \frac{f(2+x)}{f(3)} \right]^{\frac{18}{(x-1)^2}}\) is equal to:

• A. 2
• B. 1
• C. 18
• D. 9

Hint:

For limits of the form \(1^\infty\), \(\lim_{x \to a} [f(x)]^{g(x)}\), the result is \(e^{\lim_{x \to a} (f(x)-1)g(x)}\).
When dealing with complex limits involving derivatives, L'Hopital's Rule is a powerful tool. Applying it successively can simplify the problem. Taylor expansion is another excellent alternative for such problems.

Answer 1

The Correct Option is A

To solve this problem, we aim to find the limit:

\[\lim_{x \to 1} \log_e \left[ \frac{f(2+x)}{f(3)} \right]^{\frac{18}{(x-1)^2}}\]

Given the information:

• \(f(3) = 18\)
• \(f'(3) = 0\)
• \(f''(3) = 4\)

Firstly, analyze the expression \(\frac{f(2+x)}{f(3)}\) around \(x = 1\). Using Taylor's expansion for a function \(f(x)\) around \(x = 3\), we have:

\(f(2+x) = f(3) + f'(3)(2+x-3) + \frac{f''(3)}{2}(2+x-3)^2 + \cdots\)

Substituting the given values, the expansion becomes:

\[ f(2+x) = 18 + 0 \cdot (x-1) + \frac{4}{2}((x-1)^2) + \cdots = 18 + 2(x-1)^2 + \cdots \]

Thus, \(\frac{f(2+x)}{f(3)}\) simplifies to:

\[ \frac{f(2+x)}{18} = 1 + \frac{2(x-1)^2}{18} + \cdots = 1 + \frac{(x-1)^2}{9} + \cdots \]

Now, consider the logarithm expression:

\[ \log_e \left[ \frac{f(2+x)}{f(3)} \right]^{\frac{18}{(x-1)^2}} = \frac{18}{(x-1)^2} \cdot \log_e \left[ 1 + \frac{(x-1)^2}{9} + \cdots \right] \]

For small values of \(u\), \(\log_e(1+u) \approx u\). Hence,

\[ \log_e \left[ 1 + \frac{(x-1)^2}{9} \right] \approx \frac{(x-1)^2}{9} \]

Substituting this approximation, the expression simplifies to:

\[ = \frac{18}{(x-1)^2} \cdot \frac{(x-1)^2}{9} = 2 \]

Therefore, the limit is:

\[ \lim_{x \to 1} \log_e \left[ \frac{f(2+x)}{f(3)} \right]^{\frac{18}{(x-1)^2}} = 2 \]

Thus, the correct answer is 2.