Find the limit of (1+(4/3n+1))^6n.
I found it as e^6. [(1+(4/3/n+1)/(3))^6]^n and by the equation (1+1/x)^x=e,it is e^6.
Write it as $\displaystyle \left( {1 + \frac{{\frac{a}{b}}}{{n + \frac{c}{b}}}} \right)^{dn} $.
Let $\displaystyle \frac{1}{m} = \frac{{\frac{a}{b}}}{{n + \frac{c}{b}}}$ so $\displaystyle n=\tfrac{am}{b}-\tfrac{c}{b}$.
Or $\displaystyle \left( {1 + \frac{a}{{bn + c}}} \right)^{dn} = \left( {\left( {1 + \frac{1}{m}} \right)^m } \right)^{\frac{{ad}}{b}} \left( {1 + \frac{1}{m}} \right)^{ - \frac{c}{b}} $.