Let `f (x) = {{:(e ^((1)/(x ^(2)))sin ""(1)/(x), x ne0),(lamda, x =(0):}, then f '(0)`

Options

A. 1
B. `-1`
C. 0
D. Does not exist

Answer

To find \( f'(0) \) for the given function

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

we will use the definition of the derivative:

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

### Step 1: Evaluate \( f(0) \)

From the definition of \( f(x) \), we know:

\[ f(0) = \lambda \]

### Step 2: Substitute \( f(h) \) for \( h \neq 0 \)

For \( h \neq 0 \):

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

### Step 3: Set up the limit

Now, substituting into the limit:

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

### Step 4: Analyze the behavior of \( e^{-\frac{1}{h^2}} \)

As \( h \to 0 \), \( e^{-\frac{1}{h^2}} \to 0 \) because the exponent \( -\frac{1}{h^2} \) approaches negative infinity.

### Step 5: Analyze \( \sin\left(\frac{1}{h}\right) \)

The term \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1 as \( h \to 0 \). Therefore, we can bound it:

\[ -e^{-\frac{1}{h^2}} \leq e^{-\frac{1}{h^2}} \sin\left(\frac{1}{h}\right) \leq e^{-\frac{1}{h^2}} \]

### Step 6: Substitute into the limit

Now, we can rewrite the limit:

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

### Step 7: Evaluate the limit

Since \( e^{-\frac{1}{h^2}} \to 0 \):

\[ \lim_{h \to 0} \frac{e^{-\frac{1}{h^2}} \sin\left(\frac{1}{h}\right)}{h} = 0 \]

Thus, we have:

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

This limit will approach \( -\infty \) or \( +\infty \) depending on the value of \( \lambda \). For \( f'(0) \) to exist and be finite, we require \( \lambda = 0 \).

### Conclusion

Thus, if \( \lambda = 0 \):

\[ f'(0) = 0 \]

If \( \lambda \neq 0 \), \( f'(0) \) does not exist.

### Final Answer

\[ f'(0) = 0 \quad \text{if } \lambda = 0 \]