# Epsilon-Delta Proof

razored
Prove that the limit exists :$$\lim_{x\rightarrow 1} 3x -1 =2$$

Here is the work to find $$\delta$$.
$$|3x-3|<\epsilon$$ Substitution
$$3|x-1|<\epsilon$$
$$|x-1|<\frac{\epsilon}{3}$$
$$0<|x-1|<\delta$$ Appears that epsilon is equal to delta.
$$\delta = \frac{\epsilon}{3}$$
I have gotten here thus far, now what? :uhh: What I have proved? The only thing I have discovered that delta is simply equal to epsilon.
I don't understand.
I also understand that you begin the proof like this:
Given $$\epsilon> 0$$, choose $$\delta = \frac_{\epsilon}{3}$$
Now what from here as well?

Thanks beforehand.

Homework Helper
First, let me remark that you derived $\delta = \epsilon / 3$ and then proceed to say that "I have discovered that delta is simply equal to epsilon." What you have discovered, is that if you take delta to be equal to epsilon divided by three (or anything smaller will do) then the definition of limit will be satisfied.

You know the definition of the limit:
$$\lim_{x \to a} f(x) = L$$ means that: for each $\epsilon > 0$ there is a $\delta = \delta(\epsilon) > 0$ such that $|f(x) - L| < \epsilon$ whenever $|x - a| < \delta$.

In general, the proof goes like this:
Let/choose $\epsilon > 0$ (arbitrarily). Define $\delta = \cdots$ and suppose $|x - a| < \delta$. Then by some sequence of inequalities, we show that $|f(x) - L| < \epsilon$.

Now compare this to the definition and see if you understand the structure of the proof. Next remark is, that to actually give such a proof, you have to know what $\delta$ is. That's why you always start on a scrap piece of paper to calculate this value for any given value of $\epsilon$, which is what you've just done. Now all you have to do is write down the formal prove as I gave it above, substituting the correct values (basically, the proof is what you wrote down reversed).