# What is the limit of ln(2x)-ln(1+x) as x goes to infinity?

Oct 6, 2015

I found $\ln \left(2\right)$

#### Explanation:

Let us use a rule of the logs:
${\lim}_{x \to \infty} \left[\ln \left(2 x\right) - \ln \left(1 + x\right)\right] =$
${\lim}_{x \to \infty} \left[\ln \left(\frac{2 x}{1 + x}\right)\right] =$ collect $x$:
${\lim}_{x \to \infty} \left[\ln \left(\frac{2 \cancel{x}}{\cancel{x} \left(\frac{1}{x} + 1\right)}\right)\right] =$
as $x \to \infty$ then $\frac{1}{x} \to 0$
$= \ln \left(2\right)$

Oct 6, 2015

Use the properties of $\ln$ to rewrite as a single $\ln$ then use continuity of $\ln$.

#### Explanation:

Rewrite as $\ln u$, tghen use the following.

Because $\ln$ is continuous on $\left(0 , \infty\right)$, we have

if ${\lim}_{x \rightarrow \infty} u = L$ for some number $L$, then

${\lim}_{x \rightarrow \infty} \ln u = \ln \left({\lim}_{x \rightarrow \infty} u\right)$.

If you just want the answer it is $\ln 2$.