How do you use a graphing calculator to find the limit of the sequence #a_n=2^n/(2^n+1)#?

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Answer 1 · 2016-12-10T13:11:03

Plot it as a continuous function. As you are looking at end behaviour it does not matter that you are plotting it as 'every where dense'

#y=(2^x)/(2^x+1)#

Every where dense #-> # all values rather than just select values

Write as #a_n=(2^n)/(2^n(1+1/2^n)) = 1/(1+1/2^n)#

When #n=0" "# then we have #" "a_0=1/2#

#lim_(n->+oo) a_n= 1/1=1#

#lim_(n->-oo) a_n=1/(1+oo) ->0#

Tony B