# Limit Of A Function From Definition

1. Oct 20, 2009

### 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!

2. Oct 20, 2009

### 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?

3. Oct 20, 2009

### 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

4. Oct 20, 2009

### 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??

5. Oct 20, 2009

### arildno

EXACTLY!

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

6. Oct 20, 2009

### 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

7. Oct 20, 2009

### arildno

Slightly scruffy-looking proof, but okay nonetheless!

8. Oct 20, 2009

### Juggler123

Thanks for the help!