How do you use the formal definition of a limit to find #lim 4x -1# as x approaches 2?

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Answer 1 · 2018-04-26T14:24:06

#lim_(x->2) (4x-1) = 7#

Let:

#x = 2+xi#

so that #x in (2-delta,2+delta) => abs xi < delta#.

Evaluate now:

#abs (4x-1-L) = abs(4(2+xi)-1-L) = abs(4xi +7-L)#

If we let #L=7#, then:

#abs (4x-1-L) =4 abs(xi) <4 delta#

and then for any number #epsilon > 0# we can choose #delta_epsilon < epsilon/4# and then:

#x in (2-delta_epsilon,2+delta_epsilon) => abs (4x-1-L) < 4delta_epsilon < epsilon#

which proves that:

#lim_(x->2) (4x-1) = 7#