Calculate limit without L'Hopital I need a some help with this.
Calculate:
$$\lim_{x\to\infty}\frac{\ln(x-1)}x$$
I know the answer is zero. But dont know how to handle the $\ln(x-1)$
 A: Suppose $x>2$, which is not restrictive, and set $x-1=e^t$; thus $x=e^t+1$ and the limit becomes
$$
\lim_{t\to\infty}\frac{t}{e^t+1}
$$
Since
$$
0\le\frac{t}{e^t+1}\le\frac{t}{e^t}
$$
the limit follows at once from the well-known fact that
$$
\lim_{t\to\infty}\frac{e^t}{t}=\infty
$$
A: You have that $$\log \sqrt{x -1} < \sqrt{x - 1} \Rightarrow \log (x-1) < 2\sqrt{x-1} \Rightarrow \frac{\log (x-1)}{x} < 2\frac{\sqrt{x-1}}{x}$$
Can you take it from here?
A: We will prove instead that $\frac{\log x}{x} \to 0$, and leave you to deduce what you need. Suppose $x \geq e$ (here $e$ is the base of the logarithm). Then
$$\frac{\log (ex)}{ex} = \frac{\log x + 1}{ex} \leq \frac{2}{e} \frac{\log x}{x}, $$
since $e \leq x$ implies $1 \leq \log x$ (since $\log x$ is monotone). Monotonicity of $\log x$ and $x$ shows that for $e \leq x \leq e^2$ we have
$$ \frac{\log x}{x} \leq \frac{\log (e^2)}{e} = \frac{2}{e}. $$
Take now an arbitrary $x$ and divide it by $e$ enough times so that you obtain something in the range $[e,e^2]$. You can express the result by $x = e^n t$, where $e \leq t \leq e^2$ and $n \geq 0$. The above two calculations show that
$$ \frac{\log x}{x} \leq \left(\frac{2}{e}\right)^{n+1}. $$
As $x\to\infty$, the corresponding power $n$ also tends to infinity, and so the ratio $\frac{\log x}{x}$ tends to zero.
A: Because $ \ln{x-1} $ is a monotonically increasing function in its domain $]1,\infty]$:
$$ 0 < \ln{x-1} < \ln{x} \Leftrightarrow 0 < \frac{\ln{x-1}}{x} < \frac{\ln{x}}{x}$$ 
Our strategy from here is to prove that the right-hand side ($ln(x)/x$) is approaching $0$ as $x$ is approaching infinity. We will then use the Squeeze theorem to prove that the target expression is approaching 0 as well.
Let's proceed by introducing the substitution $x=e^{t}$ and then utilize a standard limit in the last step:
$$\lim_{x\to\infty} \frac{\ln{x}}{x} = \lim_{t\to\infty} \frac{\ln{e^t}}{e^{t}} = \lim_{t\to\infty} \frac{t}{e^t} = 0.$$
Using the squeeze theorem, we can now conclude that:
$$\lim_{x\to\infty} \frac{\ln{x-1}}{x} = 0.$$
A: Well, if it's infinity over infinity you can still use L'Hopital's rule.
the derivative of both is:
( 1/(x - 1 ) ) / 1
invert and multiply and then plugin infinity in for X.
The denominator is getting infinitely large and therefore we can assume it's getting closer and closer to 0. Which is why it's 0 (or at least that's how my instructor taught me)
