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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.

  • 0
    @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."2012-06-08

5 Answers 5

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

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    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 $t \to 0^+$.2012-06-08
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Asymptotics ... $\begin{align} \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\ 1 + \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 + \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\ 1 - \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 - \frac{1}{x} - \frac{1}{3 x^{3}} - \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)} &= 1 + \frac{2}{x} + \frac{2}{x^{2}} + \frac{8}{3 x^{3}} + \frac{10}{3 x^{4}} + \frac{64}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\left(\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}\right)^{x} &= \operatorname{e} ^{2} + \frac{4 \operatorname{e} ^{2}}{3 x^{2}} + \frac{20 \operatorname{e} ^{2}}{9 x^{4}} + O \Bigl(x^{(-5)}\Bigr) \end{align}$

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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.2012-06-05
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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.)

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    To number, try `\tag{1}` (or similar) in your euqations, like `$\sin x \tag{1}$`2012-06-05
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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 $\exp (y)$

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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 $x\rightarrow \infty$.

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    @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.2012-06-08