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Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right] \end{align*}$.

After some minutes around this limit I did it this way:

$\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$

So,$\space x=2^y+1$.

When $x \to +\infty$,$\space y \to +\infty$ also. By substitution:

$\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}$

$\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}} \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0 \end{align*}$

Is this correct?Are there any other easy way to find this limit?Thanks

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    Yes, it is correct. I$n$tuitively, you can see this as log x growing at a slower rate than x, hence the limit tends to 0.2012-04-05

2 Answers 2

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Using change of base for logarithms, you can write $\log_2(x-1)=\frac{\ln(x-2)}{\ln(2)}$ so we have $\underset{x\to\infty}{\lim}\frac{\ln(x-1)}{x\ln(2)}$ Notice as $x\to\infty$, we get "$\frac{\infty}{\infty}$" and so we can use L'Hospital's rule and take derivatives of the numerator and denominator to get $\underset{x\to\infty}{\lim}\dfrac{\frac{1}{x-1}}{\ln(2)}=\underset{x\to\infty}{\lim}\dfrac{1}{\ln(2)(x-1)}=0$

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    You don't need to "change the base of logarithms" if you know the derivative of $\log_a(x)$, which is $\frac{1}{x \log(a)}$. But it's just a comment ; the answer is good.2012-04-03
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I'm no expert, but in a case like this you can use Hospital's rule, which states that \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right]
= \lim_ {x \to+\infty} \left [ \frac{(\log_{2}(x-1))'}{x'}\right]

If the original limit gives $0/0$ or $\infty/\infty$

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    Can you evaluate your last limit and give a result?2012-04-05