# How do you prove the statement lim as x approaches 9 for #root4(9-x) = 0# using the epsilon and delta definition?

##### 1 Answer

See explanation.

#### Explanation:

The

the limit of a function

#f(x)# , as#x# approaches some value#c# , is#L# if, for every possible#epsilon>0# , we can find a#delta>0# that depends on, such that#epsilon# #abs(f(x)-L) < epsilon# whenever#abs(x-c) < delta# .

It's like a game. Player 1 picks an *guaranteed* to get mapped to an

If we can prove that, for every

If

#AA" "epsilon > 0" "EE" "delta>0"# such that

#abs(f(x)-L) < epsilon # when#abs(x-c) < delta# then

#lim_(x->c)f(x)=L# .

Okay, so that's a lot of mumble jumble. Let's put it to use.

We need to find a *function of*

#abs(x-c) < delta => abs(f(x)-L) < epsilon#

So we start with

In this example,

#abs(x-c) < delta => abs(x-9) < delta#

#color(white)(abs(x-c) < delta) => abs(9-x) < delta#

#color(white)(abs(x-c) < delta) => root(4)abs(9-x) < root(4)delta#

#color(white)(abs(x-c) < delta) => abs(root(4)(9-x)) < root(4)delta#

#color(white)(abs(x-c) < delta) => abs(root(4)(9-x)-0) < root(4)delta#

#color(white)(abs(x-c) < delta) => abs(f(x)-0) < root(4)delta#

Hey—looks like we may have found our connection between **If we let #root(4)delta = epsilon#, then we have**

#color(white)(abs(x-c) < delta) => abs(f(x)-0) < epsilon#

**and so we've shown that**

The last thing to do is to solve

#" "root(4)delta=epsilon#

# => delta = epsilon^4#

All the work we've done here simply means that *no matter how small* an

#lim_(x->9)root(4)(9-x)=0#

has been proven.