Let $$f : R \to R$$ be a function such that $$|f(x)| \leq x^2$$, for all $$x \in R$$. Then, at x = 0, f is:

We are given a function $$ f: \mathbb{R} \to \mathbb{R} $$ such that $$ |f(x)| \leq x^2 $$ for all real numbers $$ x $$. We need to analyze the behavior of $$ f $$ at $$ x = 0 $$, specifically checking continuity and differentiability.

First, let us check continuity at $$ x = 0 $$. For continuity, we require that $$ \lim_{x \to 0} f(x) = f(0) $$.

From the given condition, $$ |f(x)| \leq x^2 $$, which implies $$ -x^2 \leq f(x) \leq x^2 $$ for all $$ x $$. Now, consider the limits as $$ x $$ approaches 0:

$$ \lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0. $$

By the Squeeze Theorem, since $$ f(x) $$ is sandwiched between $$ -x^2 $$ and $$ x^2 $$, both of which approach 0 as $$ x \to 0 $$, we have:

$$ \lim_{x \to 0} f(x) = 0. $$

Now, what is $$ f(0) $$? Using the given condition at $$ x = 0 $$:

$$ |f(0)| \leq (0)^2 = 0. $$

Since the absolute value is non-negative and bounded above by 0, we must have $$ |f(0)| = 0 $$, which implies $$ f(0) = 0 $$. Therefore,

$$ \lim_{x \to 0} f(x) = 0 = f(0), $$

so $$ f $$ is continuous at $$ x = 0 $$.

Next, we check differentiability at $$ x = 0 $$. The derivative at 0 exists if the following limit exists:

$$ f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 0}{h} = \lim_{h \to 0} \frac{f(h)}{h}, $$

since $$ f(0) = 0 $$. We need to evaluate this limit.

Again, from $$ |f(h)| \leq h^2 $$, we have:

$$ |f(h)| \leq h^2. $$

For $$ h \neq 0 $$, we can divide both sides by $$ |h| $$ (which is positive):

$$ \frac{|f(h)|}{|h|} \leq \frac{h^2}{|h|} = |h|. $$

Since $$ \frac{|f(h)|}{|h|} = \left| \frac{f(h)}{h} \right| $$, we get:

$$ \left| \frac{f(h)}{h} \right| \leq |h|. $$

This implies:

$$ -|h| \leq \frac{f(h)}{h} \leq |h|. $$

Now, take the limit as $$ h \to 0 $$:

$$ \lim_{h \to 0} (-|h|) = 0 \quad \text{and} \quad \lim_{h \to 0} |h| = 0. $$

By the Squeeze Theorem,

$$ \lim_{h \to 0} \frac{f(h)}{h} = 0. $$

Therefore, the limit exists and equals 0, so $$ f'(0) = 0 $$. Hence, $$ f $$ is differentiable at $$ x = 0 $$.

In conclusion, at $$ x = 0 $$, the function $$ f $$ is both continuous and differentiable.

Hence, the correct answer is Option C.