**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#