How do you prove that the limit of #3x+5=35# as x approaches 10 using the precise definition of a limit?

1 Answer

First we need to state this precise definition of a limit of a function #f: X to RR#:

#lim_(x to x_0) f(x) = a in RR iff#
# iff forall epsilon > 0 exists delta > 0 forall x in X : |x-x_0| < delta => |f(x)-a| < epsilon#

From a geometrical point of view, this means that saying this limit takes the value #a# is the same thing as saying that if #x# is near #x_0# , then #f(x)# has to be near #a#, and this has to be valid no matter how small we consider the distance between #f(x)# and #a#.

For #f(x) = 3x+5#, the proof that #lim_(x to 10) f(x) = 35# is very simple.

Given #epsilon > 0#, take #delta = epsilon/3#. Then, using the triangle inequality:

#|x-10| < delta => |3x+5-35| = |3x -30| = 3 |x-10| < 3 delta = 3 epsilon/3 = epsilon iff |3x+5-35| < epsilon#

This proof can be generalized to any function #f# that fulfils the condition:

#exists k geq 0 forall x_1, x_2 in X : |f(x_2)-f(x_1)| leq k |x_2 - x_1|#

Functions of this kind are caled Lipschitz continuous functions.