# Find the limit as x approaches infinity of y=ln( 2x )-ln(1+x)?

Mar 12, 2018

$\ln 2$

#### Explanation:

We have ${\lim}_{x \rightarrow \infty} \left(\ln \left(2 x\right) - \ln \left(1 + x\right)\right)$

Since $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$, we write:

${\lim}_{x \rightarrow \infty} \ln \left(\frac{2 x}{1 + x}\right)$

According to the limit chain rule, let's say we have a function $\left(f \circ g\right)$, or $f \left(g \left(x\right)\right)$. The limit of this function, when $x$ tends to $a$, is given by $L$. To find $L$, we first find ${\lim}_{x \rightarrow a} g \left(x\right)$. The answer to this is given as $b$. Then, we find ${\lim}_{u \rightarrow b} f \left(u\right)$, where $u = g \left(x\right)$. The answer for this final limit is $L$.

Here we take $f \left(x\right) = \ln x$ and $g \left(x\right) = \frac{2 x}{1 + x}$. Also, here, $a = \infty$. First we take:

${\lim}_{x \rightarrow \infty} \frac{2 x}{1 + x}$

$2 {\lim}_{x \rightarrow \infty} \frac{x}{1 + x}$

We input $\infty$:

$2 \left(\frac{\infty}{\infty}\right)$

We see that we get an indeterminate form. According to L'Hopital's Rule, when ${\lim}_{x \rightarrow c} f \frac{x}{g} \left(x\right)$ yields an indeterminate form, such as $\frac{\infty}{\infty}$, the limit is given by ${\lim}_{x \rightarrow c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$.

We go back to ${\lim}_{x \rightarrow \infty} \frac{x}{1 + x}$. Since $\frac{d}{\mathrm{dx}} x = 1$ and $\frac{d}{\mathrm{dx}} \left(1 + x\right) = 1$, we write it as:

$2 \cdot \left(\frac{1}{1}\right)$

$2 \left(1\right)$

$= 2$

So we say that $b = 2$. We compute:

${\lim}_{u \rightarrow 2} \ln \left(u\right)$

Simply input and we get:

$\ln \left(2\right)$