[Math] Understanding how to use $\epsilon-\delta$ definition of a limit

epsilon-deltalimitsproof-writingreal-analysis

I finally understand the geometric intuition behind the $\epsilon-\delta$ definition of a limit, which is actually quite neat:

enter image description here

But I'm having trouble actually using the definition to come to a conclusion.

For (a solved) example, to prove that $\lim_{n\rightarrow \infty} \frac{3n+5}{2n+7} = \frac{3}{2}$, the following solution is given:

Proof:

Let $a_n = \frac{3n+5}{2n+7}$. Then, $\left | a_n-\frac{3}{2} \right |=\frac{11}{2(2n+7)}<\frac{3}{n}$. Given $\epsilon > 0$, choose $n_0 \in \mathbb{N}$ such that $n_0 > \frac{3}{\epsilon}$. Then, for all $n \ge n_0, \left | a_n-\frac{3}{2} \right | < \epsilon $. Therefore, $\lim a_n = \frac{3}{2}$.

I'm not quite sure I understand this proof. Namely:

  1. Where did $\frac{3}{n}$ come from?
  2. I guess we just replaced n with $\epsilon$ in the third sentence.
  3. How did the fourth sentence follow from the third, and why does that show that 3/2 is the limit?

So I tried it myself on another problem, where $a_n = \frac{2n+5}{6n-3}$, and the limit being $\frac{1}{3}$:

Let $a_n = \frac{2n+5}{6n-3}$. Then, $\left | a_n-\frac{1}{3} \right |=\left | \frac{6n-6n+15-3}{3(6n-3)} \right |= \left | \frac{12}{3(6n-3)} \right |= 4\left | \frac{1}{6n-3} \right | < \epsilon$. Dividing by 4, $\left | \frac{1}{6n-3} \right | < \frac{\epsilon}{4}$.

And that's as far as I logically get… I assume the next step is to choose an $n_0 \in \mathbb{N}$ such that $n_0 > \frac{\epsilon}{4}$, and isolating epsilon, $\epsilon < 4n_0$.

but as you can tell, I'm really lost.

Best Answer

1. You just bound in by making the denominator smaller and numerator bigger, this is a common trick in these proofs.

$$\frac{11}{2(2n+7)}<\frac{11}{2(2n)}=\frac{11}{4n}<\frac{12}{4n}=\frac3n$$

These proofs have TWO STEPS. Basically, you want to do the "prep work", then the proof. Let me write out this proof in full. The above is the PREP WORK and not actually the proof.

Proof: Let $\epsilon >0$ be given. By the Archimedian property of the natural numbers, there is some $n_0>\frac{3}{\epsilon}$. Thus if $n>n_0$, then $n>\frac{3}{\epsilon}$. Rearranging gives that $\epsilon>\frac{3}{n}$. Thus:

$$\epsilon>\frac{3}{n}=\frac{12}{4n}>\frac{11}{4n}>\frac{11}{2(2n+7)}=\left|a_n-\frac{3}{2}\right|$$

That is, for all $\epsilon>0$ there is some $n_0$ such that for all $n>n_0$. $|a_n-\frac{3}{2}|<\epsilon$. This is the definition of limit, so $\lim\limits_{n\to\infty} a_n=\frac{3}{2}$.

Related Question