# How do you find the limit of [ln(6x+10)-ln(7+3x)] as x approaches infinity?

Nov 26, 2016

Convert to a fraction and then use L'Hôpital's rule. Answer: $\ln \left(2\right)$

#### Explanation:

Given: ${\lim}_{x \to \infty} \left[\ln \left(6 x + 10\right) - \ln \left(7 + 3 x\right)\right]$

Use the identity $\ln \left(a\right) - \ln \left(b\right) = \ln \left(\frac{a}{b}\right)$:

${\lim}_{x \to \infty} \ln \left(\frac{6 x + 10}{7 + 3 x}\right) =$

$\ln \left({\lim}_{x \to \infty} \frac{6 x + 10}{7 + 3 x}\right)$

Because the above evaluated at the limit is an indeterminate form, $\frac{\infty}{\infty}$, we should use L'Hôpital's rule :

Compute the derivative of numerator:

$\frac{d \left(6 x + 10\right)}{\mathrm{dx}} = 6$

Compute the derivative of the denominator:

$\frac{d \left(7 + 3 x\right)}{\mathrm{dx}} = 3$

Make a new fraction:

$\ln \left({\lim}_{x \to \infty} \left(\frac{6}{3}\right)\right)$

This can be evaluated at $\infty$:

$\ln \left(2\right)$

Therefore, the original limit goes to the same value:

${\lim}_{x \to \infty} \left[\ln \left(6 x + 10\right) - \ln \left(7 + 3 x\right)\right] = \ln \left(2\right)$