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I'm preparing for my calculus exam and I can't solve this limit:

$$\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x$$

The limit tends to $1^\infty$, which is indeterminate. I've tried several things and I couldn't solve it.

Any idea? Thanks in advance.

Pedro
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Alejandro
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    A function can tend to some value but the limit never tends, it either is or is not. – lhf Jan 13 '12 at 12:28
  • @ljf +1, although I couldn't keep from hearing your comment in Yoda's voice, as in, "Do or not do. There is no try." – Rick Decker Jun 08 '12 at 01:22

5 Answers5

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Note that $$\tag{1}\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x=\lim_{x\rightarrow\infty}e^{\displaystyle x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}=e^{\displaystyle\lim_{x\rightarrow\infty}x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}$$ since $e^x$ is a continuous function.

Note that $$\lim_{x\rightarrow\infty}x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)= \lim_{x\rightarrow\infty}\frac{\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}{\frac{1}{x}} \cdot \left(\frac{0}{0}\right)$$ We can apply the L'Hospital rule to the previous limit. Since $$\frac{d}{dx}\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{d}{dx}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right) $$ $$=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{d}{dx}\left(\frac{2}{1-\tan(1/x)}-1\right)=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2},$$ we have $$\lim_{x\rightarrow\infty}\frac{\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}{\frac{1}{x}}= \lim_{x\rightarrow\infty}\frac{\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2}}{-\frac{1}{x^2}}$$ $$\tag{2}=\lim_{x\rightarrow\infty}\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)}{(1-\tan(1/x))^2}=2.$$

Combining $(1)$ and $(2)$, we have $\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x=e^2.$

Michael Chen
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Paul
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    using calculator seems like it tends to e^2, still this looks pretty legit. Maybe you had a mistake in the process, im gonna redo the limit using your method. – Alejandro Jan 13 '12 at 12:18
  • @Alejandro: Thanks! I see the mistake now and I correct it. See my edited answer. – Paul Jan 13 '12 at 12:25
  • $\frac{d}{dx}\left(\frac{2}{1-\tan(1/x)}-1\right)=\frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2}$. Originally I missed the $2$ on the right hand side. – Paul Jan 13 '12 at 12:27
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    Now its perfect, thanks for all :) – Alejandro Jan 13 '12 at 12:28
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    Often when mathematicians want to write $e^{xyz}$ where $xyz$ is a giant expression, they will write $\exp{xyz}$ instead. Otherwise it can be hard to see what is going on. – MJD Jun 05 '12 at 17:58
  • Nice proof, but it would be easier if, before you applied L'Hopital's Rule, you made the substitution $t= \frac{1}{x}$ and let $t \to 0^+$. – Stefan Smith Jun 08 '12 at 01:10
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Asymptotics ... $$\begin{align} \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\ 1 + \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 + \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\ 1 - \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 - \frac{1}{x} - \frac{1}{3 x^{3}} - \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)} &= 1 + \frac{2}{x} + \frac{2}{x^{2}} + \frac{8}{3 x^{3}} + \frac{10}{3 x^{4}} + \frac{64}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\left(\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}\right)^{x} &= \operatorname{e} ^{2} + \frac{4 \operatorname{e} ^{2}}{3 x^{2}} + \frac{20 \operatorname{e} ^{2}}{9 x^{4}} + O \Bigl(x^{(-5)}\Bigr) \end{align}$$

GEdgar
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  • this is a really neat solution – roo Jan 13 '12 at 15:55
  • How do you get from the second-to-last line to the last line? – Michael Lugo Jan 13 '12 at 18:51
  • Also wondering how you got the last two lines. – Tyler Hilton Jun 05 '12 at 18:32
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    $(1 + Q)^x = \exp(x \ln(1+Q)) = \exp(x (Q - Q^2/2 + \ldots))$ ... The higher-order terms are best computed by software. Note in this case your function is an even function of $x$: $$ \left( \frac{1+\tan(1/(-x))}{1-\tan(1/(-x))}\right)^{-x} = \left( \frac{1-\tan(1/x)}{1+\tan(1/x)}\right)^{-x} = \left( \frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x$$ so the terms in odd powers of $1/x$ will vanish. – Robert Israel Jun 05 '12 at 18:34
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I have an alternative solution without the use of the L'Hospital rule. Start as Paul suggested, but when in the form of

$$ \lim_{x \to \infty} x \log \left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right) $$

you can use the fact that

$$ \lim_{y \to 1} \frac{\log y}{y - 1} = 1. $$

Using this limit, the limit arithmetic and a limit of a composed function. All that helps you transform the limit above into

$$ \lim_{x \to \infty} x \left(\frac{1+\tan(1/x)}{1-\tan(1/x)} - 1\right) = \lim_{x \to \infty} x \left(\frac{2\tan(1/x)}{1-\tan(1/x)}\right) = \lim_{x \to \infty} 2 \cdot \frac{\tan{1/x}}{\frac 1x} $$ Going from the second part to the third one required yet another arithmetic to get rid of the denominator - that is obviously one, because it is continuous. The last bit can be solved using yet another known limit $$ \lim_{y \to 0} \frac{\tan y}{y} = 1 $$

So we know the limit is two, we apply the exponential function and get the result $e^2$.

Hope this helps as well.

(Sorry for the typesetting mess [no eq numbers], I have yet to learn how to work with this system.)

Ondrej
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You may let the limit as $z$ and let $y=\ln(z)$, then use L'Hospital rule to find the limits of $y$ and finally $z$ can be calculated $\exp (y)$

rschwieb
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Mathematics
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EDIT: you can write your expression as $$ \bigg(1+\frac{2\tan(1/x)}{1-\tan(1/x)}\bigg)^x \sim \bigg(1+\frac{2}{x}\bigg)^x \rightarrow e^2 $$ when $x\rightarrow \infty$.

Oo3
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  • This isn't enough. You need the stronger statement that $\tan(1/x) = 1/x + O(1/|x|^2)$ and even then it still remains to be proven that the $O(1/|x|^2)$ does not affect the value of the limit. – Qiaochu Yuan Jun 06 '12 at 00:42
  • You can make a direct substitution of equivalent functions without affecting the limit's value. – Oo3 Jun 06 '12 at 07:49
  • ... when there's no cancellation of homologous terms, but it's understood, I believe. – Oo3 Jun 06 '12 at 12:14
  • No you can't (depending on what you mean by "equivalent"). For example, $1 \sim 1 + \frac{1}{x}$ as $x \to \infty$, but it doesn't follow that $1^x \sim \left( 1 + \frac{1}{x} \right)^x$. – Qiaochu Yuan Jun 06 '12 at 14:45
  • You can't apply the principle of substitution of equivalent functions in your case: you are making a substitution with a finite limit on the base and an infinite on the exponent. Functions $f(x)$ and $g(x)$ you are going to change must be both infinite or infinitesimal. – Oo3 Jun 07 '12 at 11:32
  • I don't follow. You are also making a substitution with a finite limit in the base and an infinite limit in the exponent. – Qiaochu Yuan Jun 07 '12 at 13:21
  • NO! I made a substitution about $\tan(1/x)$ and $1/x$, that's perfectly legit! Now, please excuse me, but I don't want to go on with stuff about elementary calculus course. It is well known too that most exercises are made just using equivalence of infinite/infinitesimals in order to puzzle the reader. – Oo3 Jun 07 '12 at 13:32
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    It simply isn't (until you prove it), and nothing you've written so far constitutes an argument that it is. For example, $\lim_{x \to \infty} \left(1 + \tan(1/x) - 1/x \right)^{x^3} = e^{1/3}$ but $\lim_{x \to \infty} \left(1 + \frac{1}{x} - \frac{1}{x} \right)^{x^3} = 1$. – Qiaochu Yuan Jun 07 '12 at 13:55
  • But THE THEOREM DOES NOT APPLY TO f-g FORMS !! It applies only to $f/g$,$f\cdot g$, $f^g$ and $[1+f]^g$. – Oo3 Jun 07 '12 at 13:58
  • Maybe you should actually explicitly state whatever theorem you're implicitly quoting because I was under the impression that you wanted to conclude "I can replace $\tan(1/x)$ by $1/x$ whenever it appears in a limit" and this is just wrong. Does your theorem apply to $\lim_{x \to \infty} \left(1 + \tan (1/x) \right)^{x^3} \left( 1 + \frac{1}{x} \right)^{-x^3} = e^{1/3}$? As far as I can tell this is the same form as the problem. – Qiaochu Yuan Jun 07 '12 at 13:59
  • Listen Qiaochu, I know that you're a ~80k rep, that you (think) are a genius. But think about it: stop criticizing an answer (making downvoting stop for my answer) that is LAPALISSIAN: less than 1 minute. Trying to contradict a proof that is LAPALISSIAN : more than 1 day of useless study. My question is: is that really worth? – Oo3 Jun 07 '12 at 14:07
  • Yes. This discussion isn't taking place in a vacuum. Anyone who finds this question on Google and reads this answer is going to come away with confused ideas about what kind of manipulations are permissible in computing limits. Are you at least claiming that your theorem doesn't apply to my last example? Are you at least claiming that you have an explanation of why it is different from the problem even if you don't want to expend the energy to give it? – Qiaochu Yuan Jun 07 '12 at 14:12
  • Sorry to be picky, but you don't solve limits, you evaluate limits. – Stefan Smith Jun 08 '12 at 01:08
  • @user20520: it's the fastest and simplest way I know. I transformed the expression in the form $(1+f)^g$ where $f$ is $2/x$ (that is an equivalent infinitesimal of the function $(2\tan(1/x)/(1-\tan(1/x))$ ) and $g$ is $x$. It is worth to notice that even if the method is correct, the message has been downvoted. – Oo3 Jun 08 '12 at 02:52
  • @Oo3: If you spent time showing how to verify that the underlying idea works instead of spending it complaining that people pointed out the gaps, you'd probably have 22 more reputation by now. I really like "simple idea - how to make the simple idea work" style arguments, especially when they emphasize the simple ideas that people should be developing about a subject. –  Jun 08 '12 at 13:32