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| \leq \epsilon and delta such that 0 \leq |x-a| \leq \delta

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

|2x-4| \leq \epsilon and 0 \leq |x-2| \leq \delta

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 2|x-2|\leq\epsilon, 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 \delta = \delta(\epsilon) as a function of the \epsilon we were given.
That is, write |f(x)-l| as a function |x-a| 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| \leq \delta

then |f(x)-l|=2|x-2| \leq 2\delta

hence taking 2\delta \leq \frac{\epsilon}{2} would satisfy the problem??

[arildno]

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| \leq \delta

then |f(x)-l|=2|x-2| \leq 2\delta

hence taking 2\delta \leq \frac{\epsilon}{2} would satisfy the problem??

EXACTLY! :smile:

(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 \succ 0. Then f(x) tends to l as x tends to a if, given any \epsilon in R^{+} , there exists \delta in R^{+} such that,

|f(x)-l| \prec \epsilon whenever 0 \prec |x-a| \prec \delta

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

Hence |2x-4| \prec \epsilon

Now 2|x-2| \prec \epsilon = |x-a|\prec \delta

Hence taking \delta \leq \frac{\epsilon}{2}

Then |f(x)-l| \prec \epsilon whenever 0 \prec |x-a| \prec \delta

As required

[arildno]

Slightly scruffy-looking proof, but okay nonetheless! :smile:

[Juggler123]

Thanks for the help!