Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=$

The value of $\mathop {\lim }\limits_{n \to \infty } {\left( {e \cdot {a^2} \cdot {e^3} \cdot {a^4} \cdots {e^{n - 1}} \cdot {a^n}} \right)^{\frac{1}{{{n^2} + 1}}}}$ is equal to

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in N$ for which $\lim _{x}$ ${\rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1$ is equal to . . . . . .

$\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos 3 x}{\sin x \log (1+2 x)}=$

If $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x > 3 \end{cases}$,then: