Let `f :R to R` be a function, such that `|f(x)|le x ^(4n), n in N AA n in R` then `f (x)` is:

Options

A. discontinous at `x=0`
B. continous at `x=0`
C. non-differentiable at `x=0`
D. differentiable at `x=0`

Answer

To solve the problem, we need to analyze the function \( f(x) \) given the condition \( |f(x)| \leq x^{4n} \) for \( n \in \mathbb{N} \) and \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We have the inequality \( |f(x)| \leq x^{4n} \). This means that the function \( f(x) \) is bounded by \( x^{4n} \) and \( -x^{4n} \). Therefore, we can write:
\[ -x^{4n} \leq f(x) \leq x^{4n} \]

2. **Behavior as \( x \to 0 \)**:
As \( x \) approaches 0, both \( -x^{4n} \) and \( x^{4n} \) approach 0. Thus, by the squeeze theorem, we can conclude:
\[ \lim_{x \to 0} f(x) = 0 \]
This indicates that \( f(x) \) is continuous at \( x = 0 \).

3. **Behavior as \( x \to \infty \)**:
As \( x \) approaches infinity, both \( -x^{4n} \) and \( x^{4n} \) approach infinity. Therefore, \( f(x) \) can take on values that grow without bound as \( x \) increases.

4. **Differentiability**:
To determine if \( f(x) \) is differentiable, we need to check if there are any sharp points or discontinuities in the function. Since \( f(x) \) is bounded by \( -x^{4n} \) and \( x^{4n} \), and since it approaches 0 at \( x = 0 \), we can infer that \( f(x) \) is smooth around this point.

5. **Conclusion**:
Given that \( f(x) \) is continuous at \( x = 0 \) and does not have any sharp edges or discontinuities, we can conclude that \( f(x) \) is differentiable everywhere in \( \mathbb{R} \).

### Final Answer:
Thus, the function \( f(x) \) is differentiable everywhere in \( \mathbb{R} \).