How do you prove the statement lim as x approaches 3 for #(x/5) = 3/5# using the epsilon and delta definition?

Question

Ezra Adams · Accepted Answer

Preliminary work Recall the definition: #lim_(color(red)(xrarrc)) color(blue)f(x) = color(blue)(L)#

If and only in the event 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-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))#.

We can now put an end to our preliminary thoughts and show our proof to the rest of 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)#.

Thus, based on the meaning of limit, we deduce that

#lim_(xrarr3)(x/5) = 3/5#

Adam Adkins · Answer

undefined To prove the statement using the epsilon and delta definition, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if the distance between x and 3 is less than delta, then the distance between (x/5) and 3/5 is less than epsilon.

Let's start by setting up the inequality based on the definition:

| (x/5) - 3/5 | < epsilon

Next, we can simplify the inequality:

| (x - 3)/5 | < epsilon

Multiply both sides by 5:

| x - 3 | < 5 * epsilon

Now, we can see that if we choose delta to be 5 times epsilon, the inequality holds. Therefore, for any epsilon greater than 0, we can choose delta to be 5 times epsilon, and the statement lim as x approaches 3 for (x/5) = 3/5 is proven using the epsilon and delta definition.