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Source: 2025 Tsinghua University

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42154 posts

sqing

#1 h Apr 20, 2025, 9:57 AM

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Let $A= \lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right) .$ Find the value of $[100A] .$

This post has been edited 2 times. Last edited by sqing, Apr 20, 2025, 1:20 PM

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pco

#3 h Apr 20, 2025, 12:41 PM

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sqing wrote:

Let $A=\lim_{n\to\infty}(\frac{\pi}{4}+\frac{1}{n})^n .$ Find the value of $[100A] .$

$A=0$ and so $\boxed{\lfloor 100A\rfloor=0}$

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Quantum-Phantom

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Quantum-Phantom

#4 h Apr 20, 2025, 1:11 PM

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It should be
$A=\lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right).$

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sqing

#5 h Apr 20, 2025, 1:19 PM

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Thank you very much.

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sqing

#6 h Apr 20, 2025, 1:20 PM

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Sorry.Thank pco.

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#7 h Apr 20, 2025, 1:24 PM

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Quantum-Phantom wrote:

It should be
$A=\lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right).$

$A=\lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right) =e^{\lim_{n\to\infty}n ln\tan(\frac{\pi}{4}+\frac{1}{n})}$ let $t=\frac{1}{n}$
$e^{\lim_{n\to\infty}n ln\tan(\frac{\pi}{4}+\frac{1}{n})} = e^{\lim_{t\to 0} \frac{ln\tan(\frac{\pi}{4}+t)}{t}} = e^{\lim_{t\to 0} \frac{1}{tan(\frac{\pi}{4}+t)}(sec(\frac{\pi}{4}+t))^2}= e^{\lim_{t\to 0} \frac{1}{sin(\frac{\pi}{4}+t)cos(\frac{\pi}{4}+t)}}=e^2$ $[100A] =738$

This post has been edited 3 times. Last edited by SunnyEvan, Apr 20, 2025, 1:53 PM

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sqing

#9 h Apr 20, 2025, 1:48 PM

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Prove that $\lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right) = e^2$

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sqing

#10 h Apr 20, 2025, 1:57 PM

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sqing wrote:

Prove that $\lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right) = e^2$

https://artofproblemsolving.com/community/c7h1643781p10364775
https://math.stackexchange.com/questions/1750591/limit-of-tan-function?noredirect=1
*

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#11 h Apr 20, 2025, 2:00 PM

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sqing wrote:

sqing wrote:

Prove that $\lim_{n\to\infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right) = e^2$

https://artofproblemsolving.com/community/c7h1643781p10364775
https://math.stackexchange.com/questions/1750591/limit-of-tan-function?noredirect=1
*

Thank you very much !@Sqing . :-D

:-D

Hospital always help with lots of problems .

This post has been edited 1 time. Last edited by SunnyEvan, Apr 20, 2025, 2:02 PM

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sqing

#12 h Apr 20, 2025, 2:56 PM

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Thanks.

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