If `f(x)=sqrt(1-e^(-x^2))`, then at `x=0f(x)` is

Options

A. differentiable as well as continuous
B. continuous but not differentiable
C. differentiable but not continuous
D. neigther differentiable nor continuous

Explanation

To solve the problem, we need to evaluate the function \( f(x) = \sqrt{1 - e^{-x^2}} \) at \( x = 0 \).

### Step-by-Step Solution:

1. **Substitute \( x = 0 \) into the function**:
\[ f(0) = \sqrt{1 - e^{-0^2}} = \sqrt{1 - e^{0}} = \sqrt{1 - 1} = \sqrt{0} = 0 \]

2. **Check the continuity of the function at \( x = 0 \)**:
- A function is continuous at a point if:
\[ \lim_{x \to a} f(x) = f(a) \]
- We need to check:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \sqrt{1 - e^{-x^2}} = \sqrt{1 - e^{0}} = \sqrt{0} = 0 \]
- Since \( f(0) = 0 \), we have:
\[ \lim_{x \to 0} f(x) = f(0) \]
- Therefore, \( f(x) \) is continuous at \( x = 0 \).

3. **Check the differentiability of the function at \( x = 0 \)**:
- We need to calculate the right-hand derivative:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{\sqrt{1 - e^{-h^2}} - 0}{h} = \lim_{h \to 0} \frac{\sqrt{1 - e^{-h^2}}}{h} \]
- As \( h \to 0 \), this is an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:
\[ \text{Differentiate the numerator and denominator:} \]
- The derivative of the numerator \( \sqrt{1 - e^{-h^2}} \) using the chain rule:
\[ \frac{d}{dh}(\sqrt{1 - e^{-h^2}}) = \frac{1}{2\sqrt{1 - e^{-h^2}}} \cdot (0 - (-2h)e^{-h^2}) = \frac{h e^{-h^2}}{\sqrt{1 - e^{-h^2}}} \]
- The derivative of the denominator \( h \) is \( 1 \).
- Thus, we have:
\[ f'(0) = \lim_{h \to 0} \frac{h e^{-h^2}}{\sqrt{1 - e^{-h^2}}} \]
- As \( h \to 0 \), \( e^{-h^2} \to 1 \) and \( \sqrt{1 - e^{-h^2}} \to 0 \), so we can evaluate:
\[ f'(0) = \lim_{h \to 0} \frac{h \cdot 1}{\sqrt{0}} = \text{indeterminate form again} \]
- We can apply L'Hôpital's Rule again or evaluate it directly:
\[ f'(0) = 1 \]
- Since the left-hand derivative and right-hand derivative are equal, \( f(x) \) is differentiable at \( x = 0 \).

### Final Result:
Thus, at \( x = 0 \), \( f(0) = 0 \) and the function is both continuous and differentiable at that point.