f(x)={\begin{matrix}\frac{\mathrm{ln}(1+5x)-\mathrm{ln}(1+\alpha x)}{x} & x\neq 0 \\ 10 & x=0\end{matrix}
For f(x) to be continuous
x→0limf(x)=f(0)
x→0limxln(1+5x)−ln(1+αx)=10
Using expansion of ln(1+x)=x−2x2+3x3−...., we get
x→0limx(5x−225x2+……)−(αx−2α2x2+……)=10
i.e. 5−α=10⇒α=−5