linguistic:paraphrasing
Logarithmic functions

lim x → 0 ln (1 + x) x = 1 lim x → 0 ln (1 + a x) b x = a b lim x → 0 log c (1 + a x) b x = a b ln c {\displaystyle {\begin{array}{lcl}\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}&=&1\\\displaystyle \lim _{x\to 0}{\frac {\ln(1+ax)}{bx}}&=&\displaystyle {\frac {a}{b}}\\\displaystyle \lim _{x\to 0}{\frac {\log _{c}(1+ax)}{bx}}&=&\displaystyle {\frac {a}{b\ln c}}\end{array}}}