# Solving a $1^\infty$ indeterminate form.

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.

-
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

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

-
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 $\exp (y)$

-

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.)

-
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}

-
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 $x\rightarrow \infty$.

-
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