# How do you find the limit of ln(2x)-ln(1+x) as x approaches infinity?

Jan 26, 2017

Use the property that the difference to two logarithms is division within argument.
Use L'Hôpital's rule on argument.

#### Explanation:

Given: ${\lim}_{x \to \infty} \left(\ln \left(2 x\right) - \ln \left(x + 1\right)\right)$

Use the property that the difference to two logarithms is division within argument.

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

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

Compute the derivative of the numerator:

$\frac{d \left(2 x\right)}{\mathrm{dx}} = 2$

Compute the derivative of the denominator:

$\frac{d \left(x + 1\right)}{\mathrm{dx}} = 1$

The new expression is:

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

Therefore, the original limit is the same:

${\lim}_{x \to \infty} \left(\ln \left(2 x\right) - \ln \left(x + 1\right)\right) = \ln \left(2\right)$