How do you prove that the limit of #f(x) =2x-3# as x approaches 5 is 7 using the epsilon delta proof?

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Answer 1 · 2017-09-29T15:54:52

The function #f(x) = 2x-3# is a polynomial and as such it is continuous for every #x in RR#. Then:

#lim_(x->5) f(x) = f(5) = 2 xx 5 -3 = 7#

To prove it using the limit definition consider the value:

#abs(f(x) - 7) = abs ( 2x-3-7 ) = abs(2x-10) = 2 abs(x-5)#

For #x in (5-delta,5+delta)# with #delta > 0# then we have:

#abs(f(x) - 7) = 2 abs(x-5)< 2delta#

Given any #epsilon > 0# if we then choose #delta_epsilon < epsilon/2# we have that:

#x in (5-delta_epsilon, 5+delta_epsilon) => abs(f(x) - 7) < 2 delta_epsilon < epsilon#

which proves the limit.

Answer 2 · 2017-09-29T15:57:15

Please see below.

Given #epsilon > 0#, let #delta = epsilon/2#.

Now if #0 < abs(x-5) < delta#, then

#abs((2x-3)-7) = abs(2x-10)#

# = 2 abs(x-5)#

# < 2delta# #" "# (since #abs(x-5) < delta#)

# = 2(epsilon/2)

# = epsilon#

That is: if #0 < abs(x-5) < delta#, then #abs((2x-3)-7) < epsilon#