# Limit problem ?

## Hello, I'm in trouble to calculate the limit as n->infinity of $\ln \frac{1 + {\left(3 x\right)}^{n}}{n + 5 x}$ thank you

Apr 17, 2018

${\lim}_{n \to \infty} \ln \frac{1 + {\left(3 x\right)}^{n}}{n + 5 x} = \left\{\begin{matrix}\ln \left(3 x\right) & 3 x \ge 1 \\ 0 & 3 x < 1\end{matrix}\right.$

#### Explanation:

$\ln \frac{1 + {\left(3 x\right)}^{n}}{n + 5 x} = \ln \frac{{\left(3 x\right)}^{n} \left({\left(3 x\right)}^{-} n + 1\right)}{n + 5 x} =$

$= \frac{n}{n + 5 x} \ln \left(3 x\right) + \ln \frac{{\left(3 x\right)}^{-} n + 1}{n + 5 x}$

now for $3 x \ge 1 \Rightarrow {\lim}_{n \to \infty} \ln \frac{{\left(3 x\right)}^{-} n + 1}{n + 5 x} = 0$

and for $3 x < 1 \Rightarrow {\lim}_{n \to \infty} \ln \frac{{\left(3 x\right)}^{-} n + 1}{n + 5 x} = - \ln \left(3 x\right)$

hence

${\lim}_{n \to \infty} \ln \frac{1 + {\left(3 x\right)}^{n}}{n + 5 x} = \left\{\begin{matrix}\ln \left(3 x\right) & 3 x \ge 1 \\ 0 & 3 x < 1\end{matrix}\right.$