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Solve $\space \begin{align*} \lim_ {x \to+\infty} \left [ \frac{4 \ln(x+1)}{x}\right] \end{align*}$.

I did this way:

$$\begin{align*} \lim_ {x \to+\infty} \left [ \frac{4 \ln(x+1)}{x}\right] & = 4\lim_ {x \to+\infty} \left [\frac{1}{x} \ln(x+1) \right]= \\\\=4\lim_ {x \to+\infty} \left [ \ln(x+1)^{\frac{1}{x}}\right] &= 4 \ln \left[\lim_ {x \to+\infty}(x+1)^{\frac{1}{x}}\right] =4 \cdot \ln(1)=0 \end{align*}$$

What is the rule behind the shift that I made between the $\ln$ and the $limit$?

I'm in the high school.Thanks

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  • $\begingroup$ In TeX, if you write "ln x" it looks like this: $ln x$. But if you write "\ln x", with a backslash, it looks like this: $\ln x$. Two things are different: (1) It's not italicized, and (2) There is a space between $\ln$ and $x$. $\endgroup$ – Michael Hardy Feb 17 '12 at 22:08
  • $\begingroup$ @João: Your argument is not complete, though the answer is indeed $0$. We need to show that the limit of $(1+x)^{1/x}$ is $1$. How you do this depends on what has been proved beforehand. $\endgroup$ – André Nicolas Feb 17 '12 at 22:51
  • $\begingroup$ In order to prove the limit, I tryed to change the variable, to obtain something like the number $e$.But it seems that didn't work!However my thought was, any bigger number(as bigger you want)raised to $0$ is $1$. $\endgroup$ – João Feb 18 '12 at 11:22
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As has been mentioned, since $\ln(x)$ is continuous at $x=1$, $$ \lim_{x\to1}\ln(x)=\ln\left(\lim_{x\to1}x\right)=\ln(1)=0\tag{1} $$ However, how do you know that $$ \lim_{x \to+\infty}(x+1)^{\frac{1}{x}}=1\tag{2} $$ Usually the way one shows $(2)$ is by proving the limit you started with; that is, $$ \lim_{x\to\infty}\frac{\ln(1+x)}{x}\tag{3} $$ Let $x=e^t-1$, then $(3)$ becomes $$ \lim_{t\to\infty}\frac{t}{e^t-1}\tag{4} $$ Let $f(t)=\dfrac{t}{e^t-1}$. Then $$ \begin{align} \lim_{t\to\infty}\frac{f(t+1)}{f(t)} &=\lim_{t\to\infty}\frac{t+1}{t}\frac{e^t-1}{e^{t+1}-1}\\ &=\lim_{t\to\infty}\left(1+\frac1t\right)\left(\frac{1-e^{-t}}{e-e^{-t}}\right)\\ &=\frac1e\tag{5} \end{align} $$ Limit $(5)$ says that for some $T$, when $t\ge T$, $\dfrac{f(t+1)}{f(t)}<\dfrac12$. Therefore, since $f$ is continuous on $[T,T+1]$, there is some $C$ so that $$ f(t)<C2^{-t}\tag{6} $$ Thus, $$ 0\le\lim_{x\to\infty}\frac{\ln(1+x)}{x}=\lim_{t\to\infty}\frac{t}{e^t-1}\le\lim_{t\to\infty}C2^{-t}=0\tag{7} $$ and therefore, by the Sandwich Theorem, $\lim\limits_{x\to\infty}\frac{\ln(1+x)}{x}=0$.

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  • $\begingroup$ This seems to be the nicest proof of $(\log x)/x \to 0$ as $x\to\infty$. +1 $\endgroup$ – Paramanand Singh Sep 25 '14 at 1:32
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It is that $\ln$ is a continuous function.

However, I have a problem with your last step. You've got the indeterminate form $\infty^0$. The limit is such cases is not always $1$. One of the simplest cases where that fails is $$ \lim_{x\to\infty} \left( (x+1)^{1/\ln(x+1)} \right) = e. $$

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  • $\begingroup$ Just wondering why would applying L'hospital's rule gives 1 not zero in this case (I mean the original problem)? $\endgroup$ – NoChance Feb 17 '12 at 22:15
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    $\begingroup$ @Emmad Kareem: Certainly your suggestion works, and quickly. And if we want to avoid L'Hospital's Rule, we can let $1+x=e^y$, and show that $\lim_{y\to \infty}\frac{y}{e^y}=0$ by (for example) first showing that for positive $y$, $e^y>y^2/2$. $\endgroup$ – André Nicolas Feb 17 '12 at 23:17
  • $\begingroup$ @AndréNicolas, thanks for the explanation. $\endgroup$ – NoChance Feb 18 '12 at 7:12
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I would call it the "chain rule" or just simply "continuity," specifically because

$$\lim_{x\to a} f\big(g(x)\big) =f\left(\lim_{x\to a} \,g(x)\right) $$

whenever the limits exist and $f$ is continuous in a neighborhood of $\lim_{x\to a}g(x)$.

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    $\begingroup$ Nitpick: you need $f$ to be continuous in a neighbourhood of $\lim\limits_{x\to a} g(x)$, not $a$. $\endgroup$ – Rahul Feb 18 '12 at 2:24
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Just use l'Hopital's Rule to show that the limit is equal to the limit of $$\frac{4}{x+1},$$ which is simply zero.

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Is this attempt is correct, you can put end as colored in red

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