Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be such that \( f'' \) is continuous on \( \mathbb{R} \) and \( f(0) = 1 \), \( f'(0) = 0 \), and \( f''(0) = -1 \). Then \[ \lim_{x \to \infty} \left( f \left( \sqrt{\frac{2}{x}} \right) \right)^x \text{ is ............} \text{ (correct up to three decimal places)}. \]

Hint:

For limits involving expressions like \( \left( 1 - \frac{1}{x} \right)^x \), recall that this converges to \( \frac{1}{e} \).

Answer 1

Correct Answer: 0.35

Step 1: Analyzing the given function.
We are given that \( f(0) = 1 \), \( f'(0) = 0 \), and \( f''(0) = -1 \). Since \( f''(0) = -1 \), the function \( f(x) \) has a quadratic behavior near \( x = 0 \). Therefore, we can approximate \( f(x) \) for small \( x \) as: \[ f(x) \approx f(0) + \frac{f''(0)}{2} x^2 = 1 - \frac{x^2}{2}. \]

Step 2: Evaluate \( f \left( \sqrt{\frac{2}{x}} \right) \).
We now evaluate \( f \left( \sqrt{\frac{2}{x}} \right) \). Using the approximation for \( f(x) \), we get: \[ f \left( \sqrt{\frac{2}{x}} \right) \approx 1 - \frac{\left( \sqrt{\frac{2}{x}} \right)^2}{2} = 1 - \frac{1}{x}. \]

Step 3: Evaluate the limit.
Now, we compute the limit: \[ \lim_{x \to \infty} \left( 1 - \frac{1}{x} \right)^x. \] This is a standard limit and is known to approach \( \frac{1}{e} \). Thus, the value of the limit is \( \frac{1}{e} \approx 0.3679 \).

Step 4: Conclusion.
\[ \lim_{x \to \infty} \left( f \left( \sqrt{\frac{2}{x}} \right) \right)^x = \boxed{0.368}. \]