Let $f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$ be a function such that $f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}$. If the $\lim\limits _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta ; \alpha, \beta \in \mathbb{R}$, then $\alpha+2 \beta$ is equal to

$\lim \limits_{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}$ is equal to :

If the function

$$ f(x)=\left\{\begin{array}{l} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x<0 \\ 4, \quad x=0 \\ \frac{2}{x} \log _e\left(\frac{2+k_1 x}{2+k_2 x}\right), \quad x>0 \end{array}\right. $$

is continuous at $x=0$, then $k_1^2+k_2^2$ is equal to :

If $\lim _\limits{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{x}{1+x}\right)\right)^x=\alpha$, then the value of $\frac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha}$ equals :