Preliminary work
Recall the definition:
lim_(color(red)(xrarrc)) color(blue)f(x) = color(blue)(L)
If and only if
for every positive number color(blue)(epsilon) , there is a positive number color(red)(delta) for which the following is true:
If 0 < abs(color(red)(x-c))< color(red)(delta), then abs(color(blue)(f(x)-L))< color(blue)(epsilon),
To show lim_(color(red)(xrarr3)) color(blue)(x/5) = color(blue)(3/5) we need to show our reader that
for every positive number color(blue)(epsilon) , there is a positive number color(red)(delta) for which the following is true:
If 0 < abs(color(red)(x-3))< color(red)(delta), then abs(color(blue)(x/5-3/5))< color(blue)(epsilon),
We want to make abs(color(blue)(x/5-3/5)) smaller than color(blue)(epsilon), and we control (through color(red)(delta)) the size of abs(color(red)(x-3))
Notice that
abs(color(blue)(x/5-3/5)) = abs(color(blue)(1/5(x-3)))
= color(blue)(1/5)abs(color(blue)((x-3)))
= 1/5abs(color(red)((x-3)))
So, if we make abs(color(red)((x-3))) < 5color(blue)(epsilon), then we will have the desired result, because
abs(color(red)((x-3))) < 5color(blue)(epsilon) implies that 1/5abs(color(red)((x-3))) < 1/5(5color(blue)(epsilon)) which is equal to color(blue)(epsilon)).
Now, we are finished with our preliminary considerations and are ready to present our proof to the world.
(I'll keep the colors here to help us understand.)
Note also that saying a number is positive is the same as saying that it is > 0
Proof that lim_(color(red)(xrarr3)) color(blue)(x/5) = color(blue)(3/5)
Given color(blue)(epsilon) > 0 , choose color(red)(delta) = 5color(blue)(epsilon)
Now if 0 < abs(color(red)(x-3))< color(red)(delta), then we have:
abs(color(blue)(x/5-3/5)) = abs(color(blue)(1/5(x-3)))
= color(blue)(1/5)abs(color(blue)((x-3)))
= 1/5abs(color(red)((x-3)))
> 1/5(5color(blue)(epsilon))
= color(blue)(epsilon)
That is:
If 0 < abs(color(red)(x-3))< color(red)(delta), then abs(color(blue)(x/5-3/5))< color(blue)(epsilon).
So, by the definition of limit, we conclude that
lim_(xrarr3)(x/5) = 3/5
