6

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

StubbornAtom
  • 17,052
  • 1
    That is correct. You can also use L'Hopital's rule. – J126 Apr 03 '12 at 21:02
  • 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 $\frac{2^y}{y}\to\infty$. – Alex Becker Apr 03 '12 at 21:09
  • 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. – yoyostein Apr 05 '12 at 07:32

4 Answers4

4

Using change of base for logarithms, you can write $$\log_2(x-1)=\frac{\ln(x-1)}{\ln(2)}$$ 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 $$\underset{x\to\infty}{\lim}\dfrac{\frac{1}{x-1}}{\ln(2)}=\underset{x\to\infty}{\lim}\dfrac{1}{\ln(2)(x-1)}=0$$

J.G.
  • 115,835
Keaton
  • 1,214
  • 2
    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. – Patrick Da Silva Apr 03 '12 at 21:34
0

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 $\infty/\infty$

draks ...
  • 18,449
Sam Creamer
  • 23
0

Your way is fine, as an alternative we have that

$$ \frac{\log_{2}(x-1)}{x}= \frac{\log_{2}(x-1)}{x-1}\frac{x-1}{x}\to 0\cdot 1 =0$$

indeed eventually $\log_{2}(x-1) \le \sqrt{x-1}$, which is nothing but $n^2\le 2^n$ in disguise, and then

$$\frac{\log_{2}(x-1)}{x-1} \le \frac{\sqrt{x-1}}{x-1} \to 0$$

user
  • 154,566
0

We will prove that $$\lim_{x\to\infty}\frac{\log x}{x-1}=0.$$
Then, we can easily show $\lim_{x\to\infty}\frac{\log_2(x-1)}{x}=0$ as follows:$$\lim_{x\to\infty}\frac{\log_2(x-1)}{x}=\lim_{x\to\infty}\frac{1}{\log 2}\cdot\frac{\log(x-1)}{x}=\lim_{x\to\infty}\frac{1}{\log 2}\cdot\frac{\log(x-1)}{x-2}\cdot\frac{x-2}{x}\\=\lim_{x\to\infty}\frac{1}{\log 2}\cdot\lim_{x\to\infty}\frac{\log(x-1)}{x-2}\cdot\lim_{x\to\infty}\frac{x-2}{x}\\=\frac{1}{\log 2}\cdot 0\cdot 1=0.$$

Suppose $\epsilon$ is an arbitrary positive real number.
Suppose $\frac{2}{\epsilon}<x.$
Then $$\log x=\int_{1}^{x}\frac{1}{t}dt=\int_{1}^{\frac{2}{\epsilon}}\frac{1}{t}dt+\int_{\frac{2}{\epsilon}}^{x}\frac{1}{t}dt<\int_{1}^{\frac{2}{\epsilon}}\frac{1}{t}dt+\int_{\frac{2}{\epsilon}}^{x}\frac{\epsilon}{2}dt\\=\log\frac{2}{\epsilon}+\frac{\epsilon}{2}\left(x-\frac{2}{\epsilon}\right).$$
So, $$\frac{\log x}{x-1}=\frac{\int_{1}^{x}\frac{1}{t}dt}{\int_{1}^{x}1dt}<\frac{\log\frac{2}{\epsilon}+\frac{\epsilon}{2}\left(x-\frac{2}{\epsilon}\right)}{x-1}=\frac{\frac{\epsilon}{2}x+\log\frac{2}{\epsilon}-1}{x-1}=\frac{\frac{\epsilon}{2}+\frac{1}{x}\left(\log\frac{2}{\epsilon}-1\right)}{1-\frac{1}{x}}<\epsilon$$ holds for sufficiently large $x$.

tchappy ha
  • 8,690