Let the function $\mathrm{f}\left(\mathrm{x}\right)$ $=\left\{\begin{array}{cl}\frac{{\log }_{\mathrm{e}}(1+5\mathrm{x})-{\log }_{\mathrm{e}}(1+\mathrm{\alpha x})}{x};& \text{ if }\mathrm{x}\ne 0\\ 10,& \text{ if }\mathrm{x}=0\end{array}\right.$
be continuous at x=0. The $\alpha$is equal to: 

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{cc}\frac{\mathrm{In}(1+5\mathrm{x})-\mathrm{In}(1+\mathrm{\alpha x})}{\frac{\mathrm{x}}{10}},& ;\mathrm{x}\ne 0\\ ;\mathrm{x}=0& \end{array}\right.\\ \underset{\mathrm{\pi }\to 0}{lim} \frac{\mathrm{In}(1+5\mathrm{x})-\mathrm{In}(1+\mathrm{\alpha x})}{\mathrm{x}}=10\\ \end{array}$
Using expansion
$\begin{array}{l}\underset{\mathrm{\pi }\to 0}{lim} \frac{(5\mathrm{x}+\dots .)-(\mathrm{\alpha x}+\dots ..)}{\mathrm{x}}=10\\ 5-\mathrm{\alpha }=10\Rightarrow \mathrm{\alpha }=-5\end{array}$