If $\mathop {\lim }\limits_{x \to \infty } \left\{ {\ln \left( {{x^2} + 5x} \right) - 2\ln \left( {cx + 1} \right)} \right\} = -2$,then:

If $f(x) = \begin{cases} 1+\frac{2x}{a}, & 0 \leq x \leq 1 \\ ax, & 1 < x \leq 2 \end{cases}$,and $\lim_{x \rightarrow 1} f(x)$ exists,then the sum of the cubes of the possible values of $a$ is:

$\lim _{x}$ ${\rightarrow -a} \frac{x^7+a^7}{x+a} = 7$ $\Rightarrow a = ?$

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$,then:

If $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$ exists and is equal to $1$, then the value of $a$ is

If $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$,then $\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$