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dumbcow
 5 years ago
lim as n>infinity [n*tan(pi/n)]
dumbcow
 5 years ago
lim as n>infinity [n*tan(pi/n)]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can use the squeeze theorem here ^_^

nikvist
 5 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{n\rightarrow\infty}n\cdot\tan\frac{\pi}{n}=\lim_{n\rightarrow\infty}\frac{n}{\cot\frac{\pi}{n}}= \lim_{n\rightarrow\infty}\frac{1}{\frac{1}{\sin^2\frac{\pi}{n}}}=\lim_{n\rightarrow\infty}\sin^2\frac{\pi}{n}=0\]

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hmm interesting yeah i know its pi but can seem to show it mathematically

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer must be zero :)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0i seem to always get an indeterminate limit, even when using L'hopitals rule repeatedly

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0sstarica, graph function and you will see limit is not 0

nikvist
 5 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{n\rightarrow\infty}n\cdot\tan\frac{\pi}{n}= \lim_{m\rightarrow 0}\frac{\pi}{m}\cdot\tan{m}= \pi\cdot\lim_{m\rightarrow 0}\frac{\tan{m}}{m}= \pi\cdot\lim_{m\rightarrow 0}\frac{1}{\cos^2{m}}=\pi\]

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0ahh like u substitution to avoid chain rule...thank you very much
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