EXAMPLE $3$ Use this definition of a right$-$hand limit to prove the following limit$.$

lim$$x $0+$

$7$x

$=0$

SOLUTION $1.$ Guessing a value for $.$ Let $$ be a given positive number. Here a $=$ and L $=,$ so we want to find a number $$ such that

$$if$0<$ x $<$then$$

$0$

$<$that is$,$if$0<$ x $<$then$$

$<$

or$,$ raising both sides of the inequality to the seventh power, we get

if $0<$ x $<$ then $$ x $<$

$.$

This suggests we should choose $=$

$.$

$2.$ Showing that this $$ works. Given $>0,$ let $=$

$(17)$

$.$ If

$0<$ x $<,$

then

$7$x

$<$

$7$

$=$

$77$

$=$

so$|$

$7$x

$0|<.$

According to the definition, this shows that

limx $0+$

$7$x

$=0.$