Let f is twice differerntiable on R such that `f (0)=1, f'(0) =0 and f''(0) =-1,` then for `a in R, lim _(xtooo) (f((a)/(sqrtx)))^(x)=`

Options

A. `e ^(-e ^(2))`
B. `e ^((a^(2))/(4))`
C. `e ^(-(a^(2))/(2))`
D. `e ^(-2a ^(2))`

Answer

To solve the problem, we need to evaluate the limit:

\[ L = \lim_{x \to \infty} \left( f\left(\frac{a}{\sqrt{x}}\right) \right)^{x} \]

Given that \( f(0) = 1 \), \( f'(0) = 0 \), and \( f''(0) = -1 \), we can analyze the limit step by step.

### Step 1: Identify the indeterminate form

First, we substitute \( x \to \infty \):

\[ \frac{a}{\sqrt{x}} \to 0 \quad \text{as} \quad x \to \infty \]

Thus, we have:

\[ f\left(\frac{a}{\sqrt{x}}\right) \to f(0) = 1 \]

This gives us:

\[ L = \lim_{x \to \infty} \left( f\left(\frac{a}{\sqrt{x}}\right) \right)^{x} = 1^{\infty} \]

This is an indeterminate form.

### Step 2: Use the exponential limit transformation

We can use the property of limits for indeterminate forms:

\[ L = e^{\lim_{x \to \infty} \left( f\left(\frac{a}{\sqrt{x}}\right) - 1 \right) \cdot x} \]

### Step 3: Expand \( f\left(\frac{a}{\sqrt{x}}\right) \) using Taylor series

Since \( f \) is twice differentiable at \( 0 \), we can use the Taylor series expansion around \( 0 \):

\[ f\left(\frac{a}{\sqrt{x}}\right) = f(0) + f'(0) \cdot \frac{a}{\sqrt{x}} + \frac{f''(0)}{2} \left(\frac{a}{\sqrt{x}}\right)^2 + o\left(\left(\frac{a}{\sqrt{x}}\right)^2\right) \]

Substituting the known values:

\[ = 1 + 0 \cdot \frac{a}{\sqrt{x}} - \frac{1}{2} \cdot \frac{a^2}{x} + o\left(\frac{1}{x}\right) \]

Thus, we have:

\[ f\left(\frac{a}{\sqrt{x}}\right) - 1 \approx -\frac{a^2}{2x} \]

### Step 4: Substitute back into the limit

Now, substituting this back into our limit:

\[ L = e^{\lim_{x \to \infty} \left(-\frac{a^2}{2x}\right) \cdot x} \]

This simplifies to:

\[ L = e^{\lim_{x \to \infty} -\frac{a^2}{2}} = e^{-\frac{a^2}{2}} \]

### Final Answer

Thus, the limit is:

\[ \boxed{e^{-\frac{a^2}{2}}} \]