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This is a question from an old tutorial for a basic mathematical analysis module.

Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$

My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?

Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$,

it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,

and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.

Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.

Edit: Here's another attempt I've made.

After using addition formula for tangent, we get

$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$

so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have

$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$

I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts.

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  • $\begingroup$ The 'hence' statement is flawed reasoning. If that reasoning was valid, then $\underset{n\rightarrow\infty}{\lim}\left(1+\frac{1}{n}\right)^n$ would not be $e$. $\endgroup$ – Justin Benfield Apr 20 '16 at 3:34
  • $\begingroup$ @JustinBenfield that makes a lot of sense. I really should have seen that myself, so thanks for pointing it out! $\endgroup$ – jessica Apr 20 '16 at 3:42
  • $\begingroup$ The counterexample is obvious once you look at it, but the underlying reasons that that intuition is, in fact, wrong, is not at all obvious. One way to see why that naive intuition is flawed is to consider how $\tan^n$ diverges as $n\rightarrow\infty$ for any input slightly above $\frac{\pi}{4}$ (so the $\frac{1}{n}$ is basically 'racing' towards $0$ as the $\tan^n$ is 'racing' towards $\infty$, and the actual outcome is really controlled by the relationship between them). $\endgroup$ – Justin Benfield Apr 20 '16 at 3:50
  • $\begingroup$ You need the identity for $\tan(x+y) $ and $\tan \frac{1}{n}\sim\frac{1}{n}$ as $n$ goes to infinity. $\endgroup$ – Mhenni Benghorbal Apr 20 '16 at 3:53
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    $\begingroup$ I think I somewhat see the way to use the definition of e now! $$\tan(\frac{\pi}{4}+\frac{1}{n}) = \frac{\tan{\frac{\pi}{4}} + \tan{\frac{1}{n}}}{1-\tan\frac{\pi}{4}\tan{\frac{1}{n}}} = \frac{1+\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$ and further simplification gets the expression into a form that is similar to $(1+\frac{1}{n})$. Is this correct? $\endgroup$ – jessica Apr 20 '16 at 3:59
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Using the addition angle formula for the tangent function we can write

$$\tan^n\left(\frac{\pi}{4}+\frac1n\right)=\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n$$

Note that $\tan(x)=x+O(x^3)$. Then, we have

$$\begin{align} \left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n&=\left(\frac{1+\frac1n +O\left(\frac1{n^3}\right)}{1-\frac1n +O\left(\frac1{n^3}\right)}\right)^n\\\\ &=\left(\frac{1+\frac1n }{1-\frac1n }\right)^n\left(\frac{1+\frac{O\left(\frac1{n^3}\right)}{1+\frac1n}}{1 +\frac{O\left(\frac1{n^3}\right)}{1-\frac1n }}\right)^n\\\\ &\to \frac {e}{e^{-1}}\,\frac{1}{1}\,\,\text{as}\,\,n\to \infty\\\\ &=e^2 \end{align}$$

And we are done!

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  • $\begingroup$ You may want to explain how you achieved the separation of taylor series into the first two terms versus the rest in the step before you took the limit, nice answer! $\endgroup$ – Justin Benfield Apr 20 '16 at 4:06
  • $\begingroup$ @JustinBenfield Thank you! Much appreciative. The terms of interest were simply factored out of the numerator and denominator terms. Does that make sense? -Mark $\endgroup$ – Mark Viola Apr 20 '16 at 4:08
  • $\begingroup$ I understood it, but it took me a bit to see how. I have a new question though: How do you know that limit of the error terms is $\frac{1}{1}$ or that the limit of that product is the product of those limits (this is true iff you know a priori that both sequences converge)? $\endgroup$ – Justin Benfield Apr 20 '16 at 4:10
  • $\begingroup$ We can certainly write the product under the limit. Then we can see that the limits of the constituent terms do indeed have the limits as reported. To see that the "error" terms approach $1$, we can use a number of approaches. An easy one is to write $(1+O(n^{-3}))^n=e^{n\log(1+O(n^{-3}))}$ and note that $n\log(1+O(n^{-3}))\to 0$ since $\log(1+O(n^{-3}))=O(n^{-3})$. $\endgroup$ – Mark Viola Apr 20 '16 at 4:16
  • $\begingroup$ I'm not yet convinced of the last equality, but I buy the rest of that. $\endgroup$ – Justin Benfield Apr 20 '16 at 4:26
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If the only standard limit available is $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ then we are out of luck here. It is better to assume the following standard limits $$\lim_{x \to 0}\frac{\tan x}{x} = 1 = \lim_{x \to 0}\frac{\log(1 + x)}{x}$$ and then we can easily evaluate the desired limit. We have \begin{align} \log L &= \log\left\{\lim_{n \to \infty}\tan^{n}\left(\frac{\pi}{4} + \frac{1}{n}\right)\right\}\notag\\ &= \log\left\{\lim_{n \to \infty}\left(\frac{1 + \tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\ &= \log\left\{\lim_{n \to \infty}\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\ &= \lim_{n \to \infty}\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\text{ (via continuity of log)}\notag\\ &= \lim_{n \to \infty}n\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)\notag\\ &= \lim_{n \to \infty}n\cdot\dfrac{2\tan(1/n)}{1 - \tan(1/n)}\cdot\dfrac{\log\left(1 + \dfrac{2\tan(1/n)}{1 - \tan(1/n)}\right)}{\dfrac{2\tan(1/n)}{1 - \tan(1/n)}}\notag\\ &= 2\lim_{n \to \infty}\dfrac{\tan(1/n)}{1/n}\cdot\dfrac{1}{1 - \tan(1/n)}\cdot\lim_{x \to 0}\dfrac{\log\left(1 + x\right)}{x}\notag\\ &= 2\cdot 1\cdot 1\cdot 1\notag\\ &= 2\notag \end{align} Hence $L = e^{2}$.

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Solution

Applying $\textbf{L'Hospital's Rule}$, we have $$\begin{align*}\lim_{x \to +\infty} \left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]&=\lim_{x \to +\infty}\dfrac{\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\&=\lim_{x \to +\infty}\dfrac{1}{\sin\left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\cos \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \sin\left(\dfrac{\pi}{2}+\dfrac{2}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \cos\left(\dfrac{2}{x}\right)}\\&=2. \end{align*}$$ Hence, $$\begin{align*}\lim\limits_{n\to \infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right)&=\lim\limits_{x\to +\infty}\tan^x\left(\frac{\pi}{4}+\frac{1}{x}\right)\\&=\lim\limits_{x\to +\infty} \exp\left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\exp\left[\lim_{x \to \infty}x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\mathbb{e}^2.\end{align*}$$

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  • $\begingroup$ The question you are answering is $2$ years old and already has an accepted answer. Instead of this, you could focus your knowledge on currently open and active questions.There are many new questions and it would be much appreciated if you answered them instead. $\endgroup$ – The Integrator May 16 '18 at 14:12
  • $\begingroup$ Not a good way to farm "Revival" and "Necromancer" badges because this question is not popular enough. $\endgroup$ – user061703 May 16 '18 at 14:44

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