L'Hopital's Rule help

Question

The problem I've been given:

$$\lim_{x \to \infty}[\ln(7x)-\ln(x+7)]$$

I'm asked to solve it by applying L'Hôpital's rule. I know it's an indeterminate product problem, but I'm drawing a blank on how to convert this into a problem I can apply the rule to. Any help is greatly appreciated!

Thanks, Garren

Answer 1

First of all you have to eploit the logarithm properties in order to build a 'ratio', required by L'Hôpital's rule. $$\lim_{x\to\infty}[\ln(7x)-\ln(x+7)]=\lim_{x\to\infty}\ln\left(\frac{7x}{x+7}\right)$$ Then, since the function $\ln(\cdot)$ is continuous: $$\lim_{x\to\infty}\ln\left(\frac{7x}{x+7}\right)=\ln\left(\lim_{x\to\infty}\frac{7x}{x+7}\right)$$ In the end, the rule can be applied to the inner term, i.e. the limit of the rational function.

Answer 2

As Echan said $$\ln(7x)-\ln(7+x) = \ln\bigg(\dfrac{7x}{x+7}\bigg)$$ If we take the $\lim_{x\to\infty}$ directly we get $\ln\bigg(\dfrac{\infty}{\infty}\bigg)$ which is indeterminate. L'Hopital's Rule says: $$\lim\bigg(\dfrac{f(x)}{g(x)}\bigg) = \lim\bigg(\dfrac{f^{'}(x)}{g^{'}(x)}\bigg) \text{ s.t }f(x),g(x) = 0\text{ or }\pm\infty$$ This implies: $$\lim_{x\to\infty}\ln\bigg(\dfrac{7x}{x+7}\bigg) = \lim_{x\to\infty}\ln\bigg(\dfrac{(7x)^{'}}{(x+7)^{'}}\bigg) = \ln(7)$$