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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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    $\begingroup$ That is correct. You can also use L'Hopital's rule. $\endgroup$ – Joe Johnson 126 Apr 3 '12 at 21:02
  • $\begingroup$ Yes, that works. It might however be easier to note that $$0\leq\lim\limits_{y\to +\infty}\left[\frac{y}{2^y+1}\right]\leq\lim\limits_{y\to +\infty}\left[\frac{y}{2^y}\right]$$ and you know that the latter has limit $0$ because you know $\frac{2^y}{y}\to\infty$. $\endgroup$ – Alex Becker Apr 3 '12 at 21:09
  • $\begingroup$ Yes, it is correct. Intuitively, you can see this as log x growing at a slower rate than x, hence the limit tends to 0. $\endgroup$ – yoyostein Apr 5 '12 at 7:32
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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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    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Apr 3 '12 at 21:34
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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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    $\begingroup$ Can you evaluate your last limit and give a result? $\endgroup$ – draks ... Apr 5 '12 at 7:19

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