Let `f:R->R;f(x)={(-1)^n if x=1/(2^(2^n)), n=1,2,3.......... and 0` otherwisw then identity the correct statement (s).

Options

A. `lim_(xto0) f (x) f(2x)=0`
B. `lim _(xto0) f (x)` does not exist
C. `lim _(xto0) f (x) f (2x) =0`
D. ` lim_(xto0) f (x) (2x)` does not exist

Answer

To solve the problem, we need to analyze the function defined as follows:

\[ f(x) = \begin{cases} (-1)^n & \text{if } x = \frac{1}{2^{2^n}}, n = 1, 2, 3, \ldots \\ 0 & \text{otherwise} \end{cases} \]

We will determine the behavior of this function and identify the correct statements regarding it.

### Step 1: Identify Values of \( f(x) \)

We start by calculating \( f(x) \) for specific values of \( n \):

- For \( n = 1 \):
\[ x = \frac{1}{2^{2^1}} = \frac{1}{2^2} = \frac{1}{4} \quad \Rightarrow \quad f\left(\frac{1}{4}\right) = (-1)^1 = -1 \]

- For \( n = 2 \):
\[ x = \frac{1}{2^{2^2}} = \frac{1}{2^4} = \frac{1}{16} \quad \Rightarrow \quad f\left(\frac{1}{16}\right) = (-1)^2 = 1 \]

- For \( n = 3 \):
\[ x = \frac{1}{2^{2^3}} = \frac{1}{2^8} = \frac{1}{256} \quad \Rightarrow \quad f\left(\frac{1}{256}\right) = (-1)^3 = -1 \]

- For \( n = 4 \):
\[ x = \frac{1}{2^{2^4}} = \frac{1}{2^{16}} = \frac{1}{65536} \quad \Rightarrow \quad f\left(\frac{1}{65536}\right) = (-1)^4 = 1 \]

From this pattern, we observe that \( f(x) \) alternates between -1 and 1 for \( x = \frac{1}{2^{2^n}} \) depending on whether \( n \) is odd or even.

### Step 2: Behavior as \( n \to \infty \)

As \( n \) increases, \( x = \frac{1}{2^{2^n}} \) approaches 0. Therefore, we need to evaluate the limit:

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

For values of \( x \) approaching 0, we find that:

- \( f(x) = 0 \) for all \( x \) that are not of the form \( \frac{1}{2^{2^n}} \).
- As \( x \) approaches 0 from the positive side, \( f(x) \) can take values of either -1 or 1 depending on whether the last \( n \) was odd or even.

### Step 3: Left-Hand and Right-Hand Limits

- **Right-hand limit** as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \text{can be } -1 \text{ or } 1 \]

- **Left-hand limit** as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = 0 \]

Since the left-hand limit (0) does not equal the right-hand limit (which can be -1 or 1), we conclude that:

\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]

### Step 4: Conclusion

Based on the analysis, we can summarize the findings:

1. The function \( f(x) \) does not have a limit as \( x \to 0 \).
2. For values of \( x \) that are not of the form \( \frac{1}{2^{2^n}} \), \( f(x) = 0 \).
3. The values of \( f(x) \) oscillate between -1 and 1 for specific points.

### Correct Statement

The correct statement is that the limit does not exist.