Question

If \( f(3)=18,\; f'(3)=0 \) and \( f''(3)=4 \), then the value of \[ \lim_{x\to 1}\ln\left(\frac{f(x+2)}{f(3)}\right)^{\frac{18}{(x-1)^2}} is: \]

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(8\)

Hint:

For limits involving functions and exponents:
Use Taylor expansion near the given point
Convert powers using logarithms
Apply standard limits like \( \ln(1+u)\approx u \) for small \(u\)

Answer 1

The Correct Option is A

Given:

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

We are tasked with evaluating the following limit:

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

Step-by-Step Solution:

1. Simplifying the Expression:

The given expression involves a logarithm and an exponent. We can simplify it using logarithmic properties:

\[ \ln\left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^2}} = \frac{18}{(x-1)^2} \ln\left( \frac{f(x+2)}{f(3)} \right) \] Thus, the limit becomes: \[ \lim_{x \to 1} \frac{18}{(x-1)^2} \ln\left( \frac{f(x+2)}{f(3)} \right) \]

2. Expanding \( f(x+2) \) Around \( x = 1 \):

We will expand \( f(x+2) \) around \( x = 1 \) using a Taylor expansion. Since \( f(3) = 18 \), \( f'(3) = 0 \), and \( f''(3) = 4 \), we use the Taylor series for \( f(x+2) \) around \( x = 1 \):

\[ f(x+2) \approx f(3) + f'(3)(x-1) + \frac{f''(3)}{2}(x-1)^2 + \cdots \] Substituting the given values, we have: \[ f(x+2) \approx 18 + \frac{4}{2}(x-1)^2 = 18 + 2(x-1)^2 \] Thus: \[ \frac{f(x+2)}{f(3)} = \frac{18 + 2(x-1)^2}{18} = 1 + \frac{2(x-1)^2}{18} \] Now, take the natural logarithm of this expression: \[ \ln\left( 1 + \frac{2(x-1)^2}{18} \right) \] For small \( (x-1) \), we can use the approximation \( \ln(1 + y) \approx y \) when \( y \) is small. So: \[ \ln\left( 1 + \frac{2(x-1)^2}{18} \right) \approx \frac{2(x-1)^2}{18} \]

3. Substituting Back into the Limit:

Substitute this approximation back into the limit expression: \[ \lim_{x \to 1} \frac{18}{(x-1)^2} \cdot \frac{2(x-1)^2}{18} \] Simplifying: \[ \lim_{x \to 1} 2 = 2 \]

Final Answer: The value of the limit is \( \boxed{2} \).