$$\mathop {\lim }\limits_{x \to 0} {{1 - {{\cos }^3}4x} \over {{x^2}}}$$ is equal to

The value of k which makes $$f(x) = \left\{ {\matrix{ {\sin x,} & {x \ne 0} \cr {k,} & {x = 0} \cr } } \right.$$ continuous at x = 0, is

Consider the following for the two (02) items that follow:

$\text{Let } f(x)= \begin{cases} x^3, & x^2 < 1 \\ x^2, & x^2 \ge 1 \end{cases} \\$

What is  $\lim_{x \to 0} f'(x)$ equal to?

Consider the following for the two (02) items that follow:

$\text{Let } f(x)= \begin{cases} x^3, & x^2 < 1 \\ x^2, & x^2 \ge 1 \end{cases} \\$

Consider the following statements:

I. The function is continuous at $x=-1$.

II. The function is differentiable at $x=1$.

Which of the statements given above is/are correct?