$\mathop {\lim }\limits_{n \to \infty } \sin (\pi \sqrt {{n^2} + 1} ) = $

The value of $f(0)$ so that the function $f(x) = \frac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^n} - 1}}{x} = $

For a certain value of $c$, $\mathop {Lim}\limits_{x \to - \infty } [(x^5 + 7x^4 + 2)^c - x]$ is finite and non-zero. The value of $c$ and the value of the limit are:

$\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)=$

Evaluate the limit: $\lim _{x \rightarrow \infty}\left\{x-\sqrt[n]{\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)}\right\}$, where $a_1, a_2, \ldots, a_n$ are positive rational numbers.