Limit Of A Function From Definition

Juggler123

I have to prove from the definiton of the limit of a function that the limit of f(x) as x tends to 2 equlas 3, given that;

f(x)=(2x-1)

I know that I have to find an epsilon such that |f(x)-l| [tex]\leq[/tex] [tex]\epsilon[/tex] and delta such that 0 [tex]\leq[/tex] |x-a| [tex]\leq[/tex] [tex]\delta[/tex]

Nowing putting in the conditions for this f(x);

|2x-4| [tex]\leq[/tex] [tex]\epsilon[/tex] and 0 [tex]\leq[/tex] |x-2| [tex]\leq[/tex] [tex]\delta[/tex]

But I don't where to go from here. Any help would be great!

arildno

Well, you have:

|2x-4|=2|x-2|

Can this be made arbitrarily small (i.e, below some fixed "epsilon"), as long as we pick a small enough "delta"?

In particular, IF we want [itex]2|x-2|\leq\epsilon[/itex], what is the largest value we can pick for "delta" in order to ENSURE the validity of the above inequality?

scottie_000

Often, we need to find [tex]\delta = \delta(\epsilon)[/tex] as a function of the [tex]\epsilon[/tex] we were given.
That is, write [tex]|f(x)-l|[/tex] as a function [tex]|x-a|[/tex] and use your delta to show that it is less than epsilon

Juggler123

So I know that |f(x)-l|=2|(x-2)|=2|x-a|

Would it then be right to say that since

|x-a|=|x-2| [tex]\leq[/tex] [tex]\delta[/tex]

then |f(x)-l|=2|x-2| [tex]\leq[/tex] 2[tex]\delta[/tex]

hence taking 2[tex]\delta[/tex] [tex]\leq[/tex] [tex]\frac{\epsilon}{2}[/tex] would satisfy the problem??

arildno

EXACTLY!

(you meant, I think delta<=epsilon/2, I think..)

Juggler123

Yeah that is what I meant! Woops. So is the following proof correct then;

Definition of the Limit of a Function at a Point.
Suppose that f(x) is defined on (a-R,a)U(a,a+R), for some R [tex]\succ[/tex] 0. Then f(x) tends to l as x tends to a if, given any [tex]\epsilon[/tex] in R[tex]^{+}[/tex] , there exists [tex]\delta[/tex] in R[tex]^{+}[/tex] such that,

|f(x)-l| [tex]\prec[/tex] [tex]\epsilon[/tex] whenever 0 [tex]\prec[/tex] |x-a| [tex]\prec[/tex] [tex]\delta[/tex]

Now in this case f(x)=2x-1, l=3 and a=2

Hence |2x-4| [tex]\prec[/tex] [tex]\epsilon[/tex]

Now 2|x-2| [tex]\prec[/tex] [tex]\epsilon[/tex] = |x-a|[tex]\prec[/tex] [tex]\delta[/tex]

Hence taking [tex]\delta[/tex] [tex]\leq[/tex] [tex]\frac{\epsilon}{2}[/tex]

Then |f(x)-l| [tex]\prec[/tex] [tex]\epsilon[/tex] whenever 0 [tex]\prec[/tex] |x-a| [tex]\prec[/tex] [tex]\delta[/tex]

As required

arildno

Slightly scruffy-looking proof, but okay nonetheless!

Juggler123

Thanks for the help!