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

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

Let $L(m)$ be the $x$-coordinate of the left endpoint of the intersection of the graphs of $y = x^2 - 6$ and $y = m$,where $-6 < m < 6$. The value of $\mathop {\lim }\limits_{m \to 0} \left( {\frac{{L\left( { - m} \right) - L\left( m \right)}}{m}} \right)$ equals

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$ is equal to

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

$\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3} = $