# Thread: Epsilon Delta Proof with Infinity

1. ## Epsilon Delta Proof with Infinity

Hi,

I have a question about an Epsilon Delta proof in the form of f(x) = L, where x goes to infinity and L is 1/3.

(x^2 + x - 1) / (3*x^2 - 1) = 1/3 (where x goes to infinity)

For the question I did the following things:

I wrote the whole expression (A) in the following way:

A - (1/3) < Epsilon

The function did not get simplified nicely at the end, and I had to to use the square root formula to get x at the end:

I got this as inequality for x:

x > (1 + sqrt(1 - 8Epsilon + 12Epsilon^2)) / 6Epsilon

From that point I didn't know how to continue. I will be more than glad, if you could give me an idea.

2. Originally Posted by kaka87
Hi,

I have a question about an Epsilon Delta proof in the form of f(x) = L, where x goes to infinity and L is 1/3.

(x^2 + x - 1) / (3*x^2 - 1) = 1/3 (where x goes to infinity)

For the question I did the following things:

I wrote the whole expression (A) in the following way:

A - (1/3) < Epsilon

The function did not get simplified nicely at the end, and I had to to use the square root formula to get x at the end:

I got this as inequality for x:

x > (1 + sqrt(1 - 8Epsilon + 12Epsilon^2)) / 6Epsilon

From that point I didn't know how to continue. I will be more than glad, if you could give me an idea.
Just a simple idea.
Let $\displaystyle f(x):=\frac{x^{2}+x-1}{3x^{2}-1}$ for $\displaystyle x\geq\sqrt{1/3}$ (we may replace $\displaystyle \sqrt{1/3}$ with a larger one since we take limit as $\displaystyle x\to\infty$).
Define $\displaystyle g(x):=f(1/x)=\frac{x^{2}-x-1}{x^{2}-3}$ for $\displaystyle x\geq\sqrt{3}$.
Now, (although it is obvious) show that $\displaystyle \lim_{x\to0^{+}}g(x)=1/3$, which is equivalent to showing $\displaystyle \lim_{x\to+\infty}f(x)=1/3$

3. Thank you very much However, could you please describe it more clearly? I am new to epsilon delta proofs, so I don't know the intermediate steps very well (Considering the last part of your idea).

4. Originally Posted by kaka87
Thank you very much However, could you please describe it more clearly? I am new to epsilon delta proofs, so I don't know the intermediate steps very well (Considering the last part of your idea).
Clearly, you can see that the $\displaystyle g$ is increasing in a neighborhood of $\displaystyle 0$ (you can see it by differentiating) and $\displaystyle g(0)=1/3$. We have to show that for every $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta>0$ such that $\displaystyle |x|\leq\delta$ implies $\displaystyle |h(x)|\leq\varepsilon$, where $\displaystyle h(x):=g(x)-1/3=\frac{x(2x-3)}{3(x^{2}-3)}$ for $\displaystyle x\in\mathbb{R}$. Note also that since $\displaystyle g$ is increasing, then $\displaystyle h$ is also increasing and $\displaystyle h(0)=0$, which implies that $\displaystyle h$ is positive in a left neighborhood of $\displaystyle 0$ while it is positive in right neighborhood.
Now consider the case, $\displaystyle x\leq0$ but arbitrarily closer to $\displaystyle 0$. Then, for every $\displaystyle \varepsilon>0$, we have to find $\displaystyle \delta_{1}>0$ such that $\displaystyle -\delta_{1}\leq x\leq0$ implies $\displaystyle |v(x)|=-v(x)\leq\varepsilon$.
By considering the increasing nature of $\displaystyle v$, we solve $\displaystyle -v(-\delta_{1})=\varepsilon$ and get $\displaystyle \delta_{1}(\varepsilon):=\frac{3(-1+\sqrt{1+8\varepsilon+12\varepsilon^{2}})}{2(3\va repsilon+2)}>0$ (we picked the positive root of the quadratic equation).
Now let $\displaystyle x\geq0$ but again closer to $\displaystyle 0$.
Then, for every $\displaystyle \varepsilon>0$, we have to find $\displaystyle \delta_{2}>0$ such that $\displaystyle 0\leq x\leq\delta_{2}$ implies $\displaystyle |v(x)|=v(x)\leq\varepsilon$.
In this case, we solve $\displaystyle v(\delta_{2})=\varepsilon$ and get $\displaystyle \delta_{2}(\varepsilon):=\frac{3(-1+\sqrt{1-8\varepsilon+12\varepsilon^{2}})}{2(3\varepsilon-2)}>0$ (we picked the smaller root in a very small neighborhood of $\displaystyle 0$).
We therefore for every $\displaystyle \varepsilon>0$ have $\displaystyle |v(x)|\leq\varepsilon$ whenever $\displaystyle |x|\leq\delta(\varepsilon)$, where $\displaystyle \delta(\varepsilon):=\min\{\delta_{1}(\varepsilon) ,\delta_{2}(\varepsilon)\}=\delta_{1}(\varepsilon) >0$ for $\displaystyle \varepsilon>0$.
This completes the proof.
But I have to say that to learn $\displaystyle \varepsilon-\delta$ technique this example is not a good start...