$\lim _{n \rightarrow \infty} \frac{\left[6^2+12^2+18^2+\ldots+(6 n)^2\right]^2}{[5+10+15+\ldots+5 n]\left[2^3+4^3+6^3+\ldots+(2 n)^3\right]} =$

If $\lim_{x}$ ${\rightarrow 0} \left\{ \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4} \right) \right\} = 2^{-k}$,then the value of $k$ is

If $f(x) = \begin{cases} \frac{\sin(1+[x])}{[x]}, & \text{for } [x] \neq 0 \\ 0, & \text{for } [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer function,then $\lim_{x \rightarrow 0^{-}} f(x)$ is equal to

Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots \left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}$ for $n \geq 2$. Then,the value of $\lim _{n \rightarrow \infty} x_{n}$ is

The value of $\lim _{n}$ ${\rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]$ is

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$