What is $\lim_{x\to\infty} \ln(\frac{x}{x+1})$?

Question

I know how to solve $\lim_{x\to\infty} \ln(\frac{x + 1}{x})$:

$ln(\frac{x + 1}{x}) = \ln(\frac{x}{x}+\frac{1}{x})$

And the limit of that as $x$ goes to infinity is $\ln(1)$, which is $0$.

What about when the numerator and denominator are switched as in this problem?

$\lim_{x\to\infty} \ln(\frac{x}{x+1})$

This is of course equivalent to:

$\lim_{x\to\infty} \ln(x) - \lim_{x\to\infty} \ln(x+1)$

But that's not very informative because both are infinity, and $\infty - \infty$ is undefined.

Answer 1

Hint:

We can rewrite that as: $$ \ln\left(\cfrac{x}{x+1}\right) = \ln\left(\cfrac{1}{1+\frac{1}{x}}\right)= - \ln\left(1+\frac{1}{x}\right) $$

Are you able to finish?

Answer 2

$y = \lim_{x\to\infty} \ln(\frac{x}{x+1}) = \lim_{x\to\infty} \ln(\frac{1}{1+\frac{1}{x}})$

(Dividing Nr. and Dr. by x)

Now substituting $\infty$

$y = \lim_{x\to\infty} \ln(\frac{1}{1+\frac{1}{x}}) = \ln\frac{1}{1+0} = \ln(1) = 0$

Answer 3

Still shorter with equivalents: $$\frac x{x+1}\sim_\infty \frac xx=1,\quad\text{so }\quad \ln\Bigl(\frac x{x+1}\Bigr)\sim_\infty\ln(1)=0$$

Answer 4

Hint:

A well-known property of logarithms is

$$\ln\left(y\right)=-\ln\left(\frac1y\right),$$

so

$$\ln\left(\frac x {x+1}\right) = -\ln\left(\frac {x+1}x\right),$$

and you said you solved the limit of the expression on the right side.