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

Step 1: Understanding the Question:
We need to evaluate a limit involving a function \(f(x)\) for which we know the value, first derivative, and second derivative at a specific point. The structure of the limit suggests a \(1^\infty\) indeterminate form inside the logarithm.
Step 2: Key Formula or Approach:
First, simplify the expression using logarithm properties: \(\log(A^B) = B \log(A)\).
\[ L = \lim_{x \to 1} \frac{18}{(x-1)^2} \log_e \left[ \frac{f(2+x)}{f(3)} \right] \] As \(x \to 1\), \(2+x \to 3\), so \(f(2+x) \to f(3)\). The term inside the log approaches 1, and \(\log(1)=0\). The denominator \((x-1)^2 \to 0\). This is a \(\frac{0}{0}\) indeterminate form, which can be solved using L'Hopital's Rule.
Step 3: Detailed Explanation:
Let's apply the logarithm property: \[ L = \lim_{x \to 1} \frac{18 \log_e(f(2+x)) - 18 \log_e(f(3))}{(x-1)^2} \] This is in the \(\frac{0}{0}\) form since \(f(2+1) = f(3)\), so \(\log_e(f(3)) - \log_e(f(3)) = 0\).
We apply L'Hopital's Rule by differentiating the numerator and denominator with respect to \(x\):
Numerator derivative: \(\frac{d}{dx} [18 \log_e(f(2+x)) - 18 \log_e(f(3))] = 18 \frac{f'(2+x)}{f(2+x)} \times 1 - 0 = \frac{18f'(2+x)}{f(2+x)}\).
Denominator derivative: \(\frac{d}{dx} [(x-1)^2] = 2(x-1)\).
\[ L = \lim_{x \to 1} \frac{\frac{18f'(2+x)}{f(2+x)}}{2(x-1)} \] Substituting \(x=1\), the numerator becomes \(\frac{18f'(3)}{f(3)} = \frac{18 \times 0}{18} = 0\). The denominator is also 0. So we have a \(\frac{0}{0}\) form again.
Apply L'Hopital's Rule one more time:
Numerator derivative: \(\frac{d}{dx} \left[ \frac{18f'(2+x)}{f(2+x)} \right] = 18 \frac{f''(2+x)f(2+x) - (f'(2+x))^2}{(f(2+x))^2}\).
Denominator derivative: \(\frac{d}{dx} [2(x-1)] = 2\).
\[ L = \lim_{x \to 1} \frac{18 \frac{f''(2+x)f(2+x) - (f'(2+x))^2}{(f(2+x))^2}}{2} \] Now, substitute \(x=1\):
\[ L = \frac{18}{2} \frac{f''(3)f(3) - (f'(3))^2}{(f(3))^2} \] Substitute the given values: \(f(3)=18, f'(3)=0, f''(3)=4\).
\[ L = 9 \times \frac{4 \times 18 - (0)^2}{(18)^2} \] \[ L = 9 \times \frac{4 \times 18}{18 \times 18} = 9 \times \frac{4}{18} = \frac{36}{18} = 2 \] Alternative method using Taylor expansion:
Let \(h = x-1 \to 0\). Then \(x = 1+h\). The argument of f is \(2+x = 3+h\). By Taylor's theorem around \(t=3\): \(f(3+h) \approx f(3) + f'(3)h + \frac{f''(3)}{2!}h^2 + \dots\) \(f(3+h) \approx 18 + 0 \cdot h + \frac{4}{2}h^2 = 18 + 2h^2\). The limit becomes: \(\lim_{h \to 0} \log_e \left[ \frac{18+2h^2}{18} \right]^{\frac{18}{h^2}} = \lim_{h \to 0} \frac{18}{h^2} \log_e\left(1 + \frac{2h^2}{18}\right) = \lim_{h \to 0} \frac{18}{h^2} \log_e\left(1 + \frac{h^2}{9}\right)\). Using the standard limit \(\lim_{u \to 0} \frac{\log(1+u)}{u} = 1\). Let \(u = h^2/9\). \[ L = \lim_{h \to 0} \frac{18}{h^2} \left( \frac{\log_e(1 + h^2/9)}{h^2/9} \right) \times \frac{h^2}{9} = \lim_{h \to 0} \frac{18}{h^2} \times 1 \times \frac{h^2}{9} = \frac{18}{9} = 2 \] Step 4: Final Answer:
The value of the limit is 2.