$\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9} = $

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

The value of $\lim_{x \to 0} \frac{\log_{e}(\sec(ex) \cdot \sec(e^{2}x) \cdot ... \cdot \sec(e^{10}x))}{e^{2} - e^{2\cos x}}$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \sqrt x (\sqrt {x + 5} - \sqrt x ) = $

If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then:

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$