How do you use the epsilon delta definition to find the limit of #(2n) / (3n+1)# as x approaches #oo#?

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Answer 1 · 2016-11-13T19:14:30

You don't.

Find #lim_(xrarroo)(2x)/(3x-1)#

#lim_(xrarroo)(2x)/(3x-1) = lim_(xrarroo)(2)/(3-1/x) = 2/(3-0) = 2/3#

There is no epsilon-delta definition for limits at infinity. A common definition is:

#lim_(xrarroo)f(x) = L# #" "# if and only if

for every positive #epsilon#, there is an #M# such that for every #x>M#, we have #abs(f(x)-L) < epsilon#.

To prove that the limit is #2/3#, we need to show that the definition is satisfied.

In this case, let #M = 2/(3 epsilon)#.

For every #x > 2/(3 epsilon)#, we have #3x+1 > x > 2/(3 epsilon)#.

Therefore, for every #x > 2/(3 epsilon)#, we have #epsilon > 2/(3(3x+1))# and,

since #abs ((2x)/(3x-1)-2/3) = abs ((-2)/(3(3x+1))#, we have

#abs ((2x)/(3x-1)-2/3) < epsilon#