Verify the following limit?

Question

Use the #epsilon − delta# definition of a limit to verify the following limit:

# lim_(x→3) (x)/(6-x)= 1#

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Answer 1 · 2018-02-27T21:21:05

Evaluate the difference:

#abs(x/(6-x)-1) = abs ( ( x-6+x)/(6-x)) = abs ((2x-6)/(6-x)) = 2 abs (x-3)/abs(x-6)#

Given now #epsilon > 0# choose #delta_epsilon < min(1,epsilon)#

For #x in (3-delta_epsilon,3+delta_epsilon)# then we have:

#abs (x-3) < delta_epsilon#

#3-delta_epsilon < abs(x-6) < 3+delta_epsilon#

then as #delta_epsilon < 1#

#abs(x-6) > 2#

So for #x in (3-delta_epsilon,3+delta_epsilon)#:

#abs(x/(6-x)-1) < (2delta_epsilon)/ 2 #

#abs(x/(6-x)-1) < delta_epsilon #

and as #delta_epsilon < epsilon#:

#x in (3-delta_epsilon,3+delta_epsilon) => abs(x/(6-x)-1) < epsilon #

which proves the limit.