Let the function f(x) = \ cases _ e (1 + 5x) - _ e (1 + ) x & ifx ≠...

$f$\left(x\right) = \left\{$\begin{cases}\dfrac{\ln \left(1 + 5x\right) - \ln \left(1 + \alphax\right)}{x} & x \ne 0 \\ 10 & x = 0\end{cases}$\right)

For $f\left(x\right)$ to be continuous

$\lim_{x \to 0}f\left(x\right) = f\left(0\right)$

$\lim_{x \to 0}\dfrac{\ln \left(1 + 5x\right) - \ln \left(1 + \alphax\right)}{x} = 10$

Using expansion of $\ln \left(1 + x\right) = x - \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3} - . . . .$, we get

$\lim_{x \to 0}\dfrac{\left(5x - \dfrac{25{x}^{2}}{2} + \ldots \ldots\right) - \left(\alphax - \dfrac{{\alpha}^{2}{x}^{2}}{2} + \ldots \ldots\right)}{x} = 10$

i.e. $5 - \alpha = 10 ⇒ \alpha = - 5$