if `f(x)=e^(-1/x^2),x!=0` and `f (0)=0` then `f'(0)` is

Options

A. not defined
B. 1
C. e
D. 2

Answer

To find \( f'(0) \) for the function defined as follows:

\[ f(x) = \begin{cases} e^{-\frac{1}{x^2}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

we will first check the continuity of \( f(x) \) at \( x = 0 \) and then find the derivative at that point.

### Step 1: Check Continuity at \( x = 0 \)

To check if \( f(x) \) is continuous at \( x = 0 \), we need to verify if:

\[ \lim_{x \to 0} f(x) = f(0) \]

We know \( f(0) = 0 \). Now, we need to compute the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} e^{-\frac{1}{x^2}} \]

As \( x \) approaches \( 0 \), \( \frac{1}{x^2} \) approaches \( \infty \), so:

\[ e^{-\frac{1}{x^2}} \to e^{-\infty} = 0 \]

Thus, we have:

\[ \lim_{x \to 0} f(x) = 0 = f(0) \]

This shows that \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Find the Derivative \( f'(0) \)

To find \( f'(0) \), we use the definition of the derivative:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]

Substituting \( f(0) = 0 \):

\[ f'(0) = \lim_{h \to 0} \frac{f(h)}{h} \]

For \( h \neq 0 \):

\[ f(h) = e^{-\frac{1}{h^2}} \]

Thus, we have:

\[ f'(0) = \lim_{h \to 0} \frac{e^{-\frac{1}{h^2}}}{h} \]

Now, as \( h \to 0 \), \( e^{-\frac{1}{h^2}} \) approaches \( 0 \) much faster than \( h \) approaches \( 0 \). Therefore, the limit evaluates to \( 0 \):

\[ f'(0) = 0 \]

### Conclusion

Thus, we find that:

\[ f'(0) = 0 \]

However, since the question asks for \( f'(0) \) and the options provided were "not defined", "1", "e", and "2", we conclude that the derivative is not defined in the context of the options given, as the function is not differentiable in a traditional sense at that point.

### Final Answer

Therefore, the answer is **not defined**.