$\mathop {\lim }\limits_{m \to \infty } {\left( {\cos \frac{x}{m}} \right)^m} = $

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to

$\mathop {\lim }\limits_{x \to \pi /2} \frac{\tan 3x}{x} = $

$\lim _{x \rightarrow 3 / 2} \frac{\left(4 x^2-6 x\right)\left(4 x^2+6 x+9\right)}{\sqrt[3]{2 x}-\sqrt[3]{3}}=$

If $\lim_{x \to 0} \frac{45^x - 9^x - 5^x + 1}{(k^x - 1)(3^x - 1)} = 2$, then the value of $k$ is...

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