limit of a sequence with terms a_n = tan[(2n pi)/(1 + 8n)

[maeveoneill]

how would i go about finding the limit of

tan[(2n pi)/(1 + 8n)?

I believe this should involve the theorm that if lim n approaching inifintiy of an(series) = L and the function f is continuous at L, then lim n appraching inifity f(an) = f(L)

so..

=tan (lim n-->infinity (2n pi/ 1+ 8n)

 

[pka]

The answer depends on where one puts the droped "]".
\(\displaystyle \tan \left[ {2\pi n} \right] = 0\) for every n.

On the other hand
\(\displaystyle \left( {\frac{{2n\pi }}{{1 + 8n}}} \right) \to \frac{\pi }{4}\)

 

[Deleted member 4993]

Re: limit of a sequence

how would i go about finding the limit of

tan[(2n pi)/(1 + 8n)?

I believe this should involve the theorm that if lim n approaching inifintiy of an(series) = L and the function f is continuous at L, then lim n appraching inifity f(an) = f(L)

so..

=tan (lim n-->infinity (2n pi/ 1+ 8n)

2n/(1+8n) = 1/[{1/(2n)} + 4]

Now take the limit for n -> inf.

 

[maeveoneill]

whoops.

should be tan[(2n pi)/(1 + 8n)]

therefore.. would the answer be tan(pi/4)... cause the back of my book says the answer is 1.. but that is equivalent to .13 or something

 

[Deleted member 4993]

whoops.

should be tan[(2n pi)/(1 + 8n)]

therefore.. would the answer be tan(pi/4)... cause the back of my book says the answer is 1.. but that is equivalent to .13 or something

What is equivalent to .13 or something?

 

[maeveoneill]

whoops.

should be tan[(2n pi)/(1 + 8n)]

therefore.. would the answer be tan(pi/4)... cause the back of my book says the answer is 1.. but that is equivalent to .13 or something

What is equivalent to .13 or something?

tan (pi/4).... isnt that .13 or something??

 

[galactus]

\(\displaystyle \L\\tan(\frac{\pi}{4})=1\)