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Compute $$\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$$ without using L'Hospital's rule.

By using L'Hospital's rule and

$$\tan'( \Diamond )=( \Diamond )'(1+\tan^{2}( \Diamond ))$$ I mean by $\Diamond $ a function so I got \begin{align} \lim_{n\to +\infty}n\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right) &=\lim_{n\to +\infty}\dfrac{\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)}{\dfrac{1}{n}}\\ &=1+\tan^{2}\left(\dfrac{\pi}{3}\right)=1+\sqrt{3}^{2}=1+3=4 \end{align}

I'm interested in more ways of computing limit for this sequence.

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  • $\begingroup$ $\tan \pi/3=\sqrt3$. $\endgroup$ Sep 23, 2015 at 10:09
  • $\begingroup$ Could use the Taylor series: $\tan\left( \frac{\pi}{3} + x \right) = \sqrt{3} + 4x + O(x^2)$. $\endgroup$
    – Santeri
    Sep 23, 2015 at 10:10

6 Answers 6

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The expression is a derivative. Since $\sqrt{3} = \tan\left(\frac{\pi}{3}\right)$, the limit is

$$\lim_{n \to \infty} n \left( \tan\left(\frac{\pi}{3} + \frac{1}{n} \right) - \tan\left(\frac{\pi}{3}\right)\right)$$

That is by the substitution $h = \frac{1}{n}$, $$\lim_{h \to 0} \frac{1}{h} \left(\tan\left(\frac{\pi}{3} + h \right) - \tan \left(\frac{\pi}{3} \right)\right)$$

That is, $$\tan'(\pi/3)$$

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  • $\begingroup$ this is nice i didn't notice that $\endgroup$
    – Educ
    Sep 23, 2015 at 10:11
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$$\begin{align}\lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\sqrt 3\right)&=\lim_{n\to\infty}n\left(\frac{\sqrt 3+\tan\frac 1n}{1-\sqrt 3\tan\frac 1n}-\sqrt 3\right)\\&=\lim_{n\to\infty}\frac{4n\tan\frac 1n}{1-\sqrt 3\tan\frac 1n}\\&=\lim_{n\to\infty}\frac{4\tan\frac 1n/(1/n)}{1-\sqrt 3\tan\frac 1n}\\&=\frac{4\cdot 1}{1-0}\end{align}$$

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  • $\begingroup$ that's good work $\endgroup$
    – Educ
    Sep 23, 2015 at 10:24
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$$\lim_{n\to \infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3\right)$$ $$=\lim_{n\to \infty}\frac{\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3}{\frac{1}{n}}$$ $$=\lim_{n\to \infty}\frac{\frac{\tan\frac{\pi}{3}+\tan \frac{1}{n}}{1-\tan \frac{\pi}{3}\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$

$$=\lim_{n\to \infty}\frac{\frac{\sqrt 3+\tan \frac{1}{n}}{1-\sqrt 3\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$ $$=\lim_{n\to \infty}\frac{4\tan\frac{1}{n}}{\frac{1}{n}(1-\sqrt 3\tan\frac{1}{n})}$$ $$=4\lim_{n\to \infty}\frac{\tan\frac{1}{n}}{\frac{1}{n}}\times \lim_{n\to \infty}\frac{1}{1-\sqrt 3\tan \frac{1}{n}}$$ $$=4(1)\left(\frac{1}{1-0}\right)=\color{red}{4}$$

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  • $\begingroup$ your main idea is $$\tan(x+y)=\dfrac{(tan(x)+tan(y))}{(1-tan(x)\cdot\tan(y))}\quad \lim_{x\to 0}\dfrac{\tan}{x}=1$$ $\endgroup$
    – Educ
    Sep 23, 2015 at 10:25
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    $\begingroup$ yes, you are right. this trig formula helps separating limits $\endgroup$ Sep 23, 2015 at 10:26
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Taylor expansion would be a good idea. Around $x=0$ $$\tan(a+x)=\tan (a)+x \left(\tan ^2(a)+1\right)+x^2 \left(\tan ^3(a)+\tan (a)\right)+O\left(x^3\right)$$ Replace $x$ by $\frac 1n$ and get $$\tan(a+\frac 1n)-\tan (a)=\left(\tan ^2(a)+1\right)\frac 1n+\cdots$$

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  • $\begingroup$ could you give more details please $\endgroup$
    – Educ
    Sep 23, 2015 at 10:13
  • $\begingroup$ The expression you are working is $\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\tan\left(\dfrac{\pi}{3} \right)$ $\endgroup$ Sep 23, 2015 at 10:34
  • $\begingroup$ $$\tan(a+\frac 1n)-\tan (a)=\left(\tan ^2(a)+1\right)\frac 1n+\cdots$$ $$\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\tan\left(\dfrac{\pi}{3} \right)=\left(\tan ^2(\dfrac{\pi}{3})+1\right)\frac 1n+\cdots$$ $$\lim_{n\to+\infty}n\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\tan\left(\dfrac{\pi}{3} \right)=\left(\tan ^2(\dfrac{\pi}{3})+1\right)$$ $\endgroup$
    – Educ
    Sep 23, 2015 at 11:03
  • $\begingroup$ What else did I mean ? $\endgroup$ Sep 23, 2015 at 11:12
  • $\begingroup$ is there any difference if i use small o instead of big O $\endgroup$
    – Educ
    Sep 23, 2015 at 11:13
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\begin{align} \lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\tan\frac{\pi}{3} \right)&=\lim_{n\to\infty}n\left(\tan(\frac{\pi}{3}+\frac 1n-\frac{\pi}{3})\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big) \right)\\ &=\lim_{n\to\infty}\frac{n\sin\frac 1n}{\cos \frac 1n}\times \lim_{n\to\infty}\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big)\\ &=\frac{\lim_{n\to\infty}n\sin\frac 1n}{\lim_{n\to\infty}\cos \frac 1n}\times \Big(1+\lim_{n\to\infty}\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big)\\ &=\frac{1}{1}\times(1+\sqrt 3\sqrt 3)\\ &=4 \end{align}

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  • $\begingroup$ your main idea $$\tan(x+y)=\dfrac{(tan(x)+tan(y))}{(1-tan(x)\cdot\tan(y))}$$ with $x=\dfrac{\pi}{3}+\dfrac{1}{n}$ and $y=-\dfrac{\pi}{3}$ and $$\lim_{x\to 0}\dfrac{\sin(x)}{x}=\lim_{x\to 0}\dfrac{\cos(x)}{x}=1$$ $\endgroup$
    – Educ
    Sep 23, 2015 at 11:31
  • $\begingroup$ thanks for adding details :-) $\endgroup$
    – Math-fun
    Sep 23, 2015 at 11:35
  • $\begingroup$ ... and the end of your comment we need $\lim_{x\to0}\cos x =1$. $\endgroup$
    – Math-fun
    Sep 23, 2015 at 11:36
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It's very tempting to use L'Hopital's rule, indeed. An alternative proof may be like this: we construct approximations to the function $f (n) = \tan (\pi/3 + 1/n)$. For example, Taylor Series Approximation gives $$\tan \left (\frac {\pi} {3} + \frac {1} {n} \right) = \tan \frac {\pi} {3} + \frac {4} {n} + o \left (\frac {1} {n}\right).$$ Putting this back in the limit we have $$\lim_{n \to \infty} n \left(\tan \left(\frac {\pi} {3} + \frac {1} {n} \right) - \sqrt{3} \right) = \lim_{n \to \infty} (4 + o (1)) = 4.$$

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