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

Question

Charles Anctil · Accepted Answer

#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#

Emilia Aguilar · Answer

undefined To use the formal definition of a limit to find lim 4x - 1 as x approaches 2, we need to evaluate the expression 4x - 1 as x gets arbitrarily close to 2.

According to the formal definition of a limit, for any positive value ε (epsilon), there exists a positive value δ (delta) such that if 0 < |x - 2| < δ, then |(4x - 1) - L| < ε, where L is the limit we are trying to find.

To find the limit, we substitute the value of x into the expression 4x - 1 and simplify it. In this case, as x approaches 2, we substitute x = 2 into the expression:

lim 4x - 1 as x approaches 2 = 4(2) - 1 = 8 - 1 = 7.

Therefore, the limit of 4x - 1 as x approaches 2 is 7.