Zad. 5.32
\lim\limits_{x\to0}\frac{\sqrt[3]{1+mx}-1}{x}
1+mx=t^3
dla x=0, t=1
to
x=\frac{t^3-1}{m}
\lim\limits_{x\to0}\frac {t-1}{\frac {t^3-1}{m}}=\lim\limits_{t\to1}\frac{m(t-1)}{t^3-1}=
=\lim\limits_{t\to1}\frac{m(t-1)}{(t-1)(t^2+t+1)}=
\lim\limits_{t\to1}\frac{m}{t^2+t+1}=\frac{m}{3}