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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 $\ln(x-1)$

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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?

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    $\begingroup$ In some sense, beautiful... $\endgroup$ – ShakesBeer Nov 15 '14 at 17:09
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    $\begingroup$ $\ln$ is just a symbol used to describe a special logarithm to the base $e$. $\endgroup$ – Aaron Maroja Nov 15 '14 at 17:34
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    $\begingroup$ Not a rule, it's about notation. Also you may notice that we haven't made use of the base. $\endgroup$ – Aaron Maroja Nov 15 '14 at 17:37
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    $\begingroup$ Actually this is how you prove that $\lim_{x\to\infty} \frac{\log x}{x} = 0$, it's a general approach. I don't understand your concerns. Don't forget to use the squeeze theorem. $\endgroup$ – Aaron Maroja Nov 15 '14 at 17:50
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    $\begingroup$ why the downvote? $\endgroup$ – Aaron Maroja Jan 22 '15 at 20:57
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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 $$ \lim_{t\to\infty}\frac{e^t}{t}=\infty $$

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  • $\begingroup$ Yeah ok this may work.. The basic way to do it is to break out the fastest term and to shorten the equation. But this in a way is similar.. Thanks! $\endgroup$ – Ouizzo Nov 15 '14 at 17:26
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    $\begingroup$ well if you assume that is known that $\frac{e^t}{t} \to \infty$ you may just as well assume that is known that $\frac{\ln x}{x} \to 0$ $\endgroup$ – Ant Nov 15 '14 at 21:19
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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.

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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:

$$\lim_{x\to\infty} \frac{\ln{x-1}}{x} = 0.$$

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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)

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    $\begingroup$ You didn't read the question properly. It is said, "calculate limit without L'Hopital" $\endgroup$ – Venus Nov 15 '14 at 20:39

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