# What is the limit of ln(1+5x) - ln(3+4x) as x approaches oo?

Sep 5, 2016

$= \ln \left(\frac{5}{4}\right)$

#### Explanation:

${\lim}_{x \to \infty} \ln \left(1 + 5 x\right) - \ln \left(3 + 4 x\right)$

$= {\lim}_{x \to \infty} \ln \left(\frac{1 + 5 x}{3 + 4 x}\right)$

$= {\lim}_{x \to \infty} \ln \left(\frac{\frac{1}{x} + 5}{\frac{3}{x} + 4}\right)$

$= \ln \left(\frac{{\lim}_{x \to \infty} \frac{1}{x} + 5}{{\lim}_{x \to \infty} \frac{3}{x} + 4}\right)$

$= \ln \left(\frac{5}{4}\right)$

Sep 5, 2016

$\therefore \text{ The Reqd. Limit} = \ln \left(\frac{5}{4}\right)$.

#### Explanation:

The Reqd. Limit$= {\lim}_{x \rightarrow \infty} \left\{\ln \left(1 + 5 x\right) - \ln \left(3 + 4 x\right)\right\}$

$= {\lim}_{x \rightarrow \infty} \ln \left\{\frac{1 + 5 x}{3 + 4 x}\right\}$

$= {\lim}_{x \rightarrow \infty} \ln \left\{\frac{\cancel{x} \left(5 + \frac{1}{x}\right)}{\cancel{x} \left(4 + \frac{3}{x}\right)}\right\}$

$= {\lim}_{x \rightarrow \infty} \ln \left\{\frac{5 + \frac{1}{x}}{4 + \frac{3}{x}}\right\}$

As $\ln$ is a continuous fun., The Reqd. Limit

$= \ln \left\{{\lim}_{x \rightarrow \infty} \frac{5 + \frac{1}{x}}{4 + \frac{3}{x}}\right\}$

$= \ln \left(\frac{5 + 0}{4 + 0}\right) = \ln \left(\frac{5}{4}\right)$

$\therefore \text{ The Reqd. Limit} = \ln \left(\frac{5}{4}\right)$.