Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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.

share|cite|improve this question
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 8 '12 at 1:22
up vote 11 down vote accepted

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

share|cite|improve this answer
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
Now its perfect, thanks for all :) – Alejandro Jan 13 '12 at 12:28
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 5 '12 at 17:58

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 $\exp (y)$

share|cite|improve this answer

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.)

share|cite|improve this answer
To number, try \tag{1} (or similar) in your euqations, like $$\sin x \tag{1}$$ – Antonio Vargas Jun 5 '12 at 18:40

Asymptotics ... $$\begin{align} \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\ 1 + \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 + \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\ 1 - \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 - \frac{1}{x} - \frac{1}{3 x^{3}} - \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)} &= 1 + \frac{2}{x} + \frac{2}{x^{2}} + \frac{8}{3 x^{3}} + \frac{10}{3 x^{4}} + \frac{64}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\left(\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}\right)^{x} &= \operatorname{e} ^{2} + \frac{4 \operatorname{e} ^{2}}{3 x^{2}} + \frac{20 \operatorname{e} ^{2}}{9 x^{4}} + O \Bigl(x^{(-5)}\Bigr) \end{align}$$

share|cite|improve this answer
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 5 '12 at 18:32
$(1 + Q)^x = \exp(x \ln(1+Q)) = \exp(x (Q - Q^2/2 + \ldots))$ ... 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 5 '12 at 18:34

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 $x\rightarrow \infty$.

share|cite|improve this answer
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 6 '12 at 0:42
You can make a direct substitution of equivalent functions without affecting the limit's value. – Oo3 Jun 6 '12 at 7:49
... when there's no cancellation of homologous terms, but it's understood, I believe. – Oo3 Jun 6 '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 6 '12 at 14:45
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 7 '12 at 13:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.