If $\lim _{n \rightarrow \infty} x^n \log _e x=0$,then $\log _x 12=$

The $\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}$ equals

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{{y^2}}}{x}$,where ${y^2} = ax + b{x^2} + c{x^3}$.

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:

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

The value of $\lim _{x \rightarrow 0} \left( \frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}} \right)$ is