Epsilon-N proof with a subtraction in the denominator

Question

I'm in an introductory-level analysis class and am really struggling to grasp epsilon-N proofs, the way our professor has taught them (as opposed to simpler versions I've learned in the past) just makes no sense to me and does not seem to apply outside of a very specific subset of problems.

Essentially the only way I know how to solve is when we are proving the limit of a fraction with a denominator that is either a single term or two terms being added. Now of course a lot of the homework problems contain expressions that are the opposite.

In particular I am struggling with proving the limit of $\frac{n+5}{n^3 - 5}$. The solutions she posted do not make sense and I don't understand how she arrived there. How would you write a proof of the limit of this expression?

Answer 1

These proofs can be extremely hard in the beginning but in the end often boil down to - what I like to call - inequality acrobatics!

Most of the times, what you need to find is an expression that is greater than your expression and whose limit you know very well. This is done by changing your given expression to something that holds for larger $N$. What do I mean by that?

Take the example of the inequality, $N^2 > 10N + 20$. You can quickly check by hand and say that the inequality does not really hold for values like $N = 5$ and $N = 10$. In this case, such an inequality holds for $N > 11$ and that is sufficient for us. As $N\rightarrow \infty$, we can ignore the small values and concentrate on what happens after a point.

Returning to your example, the rule of thumb is: if the numerator has a smaller degree than the denominator, then the limit is zero. Thus, we need to prove for all $\varepsilon > 0$, we can pick an $N_0$ such that $$|\frac{N+5}{N^3 - 5}| < \varepsilon$$ for all $N > N_0$.

Now, let us try to figure out what this elusive $N_0$ is.

Notice that for $N > 2$, our fraction is positive! So, we can safely remove the $|\cdot|$ around it if we take our $N_0 > 2$. So, let's do that! Let $N_0 = 3$.

First of all, we look at the inequality $$2N > N + 5$$ which is valid for all $N > 5$. Thus, we have $$\frac{N+5}{N^3 - 5} < \frac{2N}{N^3 - 5}.$$ We can take our $N_0 = 6$ and our inequality would hold.

As we need a bigger fraction, we should try decreasing the denominator. I leave it as an exercise for you that for $N > 3$ (already satisfied by our choice of $N_0 = 6$), the following inequality holds $$\frac{N^3}{2} < N^3 - 5$$

This gives us the string of inequalities that goes $$\frac{N+5}{N^3 - 5} < \frac{2N}{N^3 - 5} < \frac{4}{N^2}$$

The last term is a simplification of $\frac{2N}{N^3/2}$

An exercise, as simple as before, is to see that $$\frac{4}{N^2} < \frac{1}{N}$$ holds for $N > 4$.

After all that strenuous activity, we have the train of inequalities $$\frac{N+5}{N^3 - 5} < \frac{2N}{N^3 - 5} < \frac{4}{N^2} < \frac{1}{N}$$ all of which are valid for $N>6$

And for the grand finale, we need to show, that for a given $\varepsilon >0$, we can find an $N_0$ such that for $N > N_0$, we have $\frac{1}{N} < \varepsilon$. It is indeed true! It even has a name! Basically, you invert $\varepsilon$ and find the next highest integer. Either this or 6 will be the value of our $N_0$. Formally, we write

Given an $\varepsilon >0$, define $N_0 = \max\{6, \frac{1}{\lfloor\varepsilon\rfloor} + 1\}$. Then, we have $$|\frac{N+5}{N^3-5}| < \varepsilon$$

Hope this helps you in understanding the method. The main idea is to remove the "sums" of the form $N^k+a$ and replace them by products of the form $bN^k$ by using inequalities which are valid for some $N > N_0$.

Answer 2

Here is how I would approach this problem. I will give the proof first, then break it down after.

Proof: Let $\varepsilon>0$. By the Archimedean property, there exists an $N\in\mathbb{N}$ such that $1/N<\varepsilon/6$. Let $n\geq\max\{N,3\}$. Since $n\geq 3$, then we have $n^2\leq n^3-5$ and $(n+5)/(n^3-5)>0$. Now, $$\begin{align*} \frac{n+5}{n^3-5} &\leq \frac{n+5}{n^2}\\ &=\frac{1}{n}+\frac{5}{n^2}\\ &\leq\frac{1}{n}+\frac{5}{n}\\ &=\frac{6}{n}\\ &\leq\frac{6}{N}<\varepsilon\end{align*}$$ which implies $$\lim\limits_{n\to\infty}\frac{n+5}{n^3-5}=0$$ and completes the proof.

Here is the break down: All these $\varepsilon-N$ or $\varepsilon-\delta$ proofs come down to are bounding things in the right way. Whenever you want to get an upper bound on a term with an expression in the denominator, you typically want to find a lower bound for the expression in the denominator. We do that three times in this proof. The first place we do this is when we use the lower bound $n^2\leq n^3-5$ for $n\geq 3$ (you can check this inductively, graphically, or algebraically if you factor $n^3-5$). This specific choice of lower bound allows us to say $$\frac{1}{n^3-5}\leq\frac{1}{n^2}$$ so that $$\frac{n+5}{n^3-5}\leq \frac{n+5}{n^2}$$ The way I found this lower bound was just by guess and check. I first wanted to choose the simplest lower bound, which was $n\leq n^3-5$, which happens to hold for $n\geq 3$. This lower bound is however insufficient. Although we do have $$\frac{n+5}{n^3-5}\leq\frac{n+5}{n}=1+\frac{5}{n}$$ the smallest bound on $1+\frac{5}{n}$ is by $1+\varepsilon$ for sufficiently large $n$, but I needed the $1$ to vanish as $n$ gets large. I then tried the next most simple choice of a lower bound for $n^3-5$, which was $n^2$, and worked as shown above.

The second place where we found an upper bound for a term with an expression in the denominator is when we observed that $5/n^2\leq 5/n$. This is because $n\leq n^2$ for all $n\in\mathbb{N}$.

The third time we bounded a term in this manner is when we said $6/n\leq 6/N$, which is simply because $N\leq n$ in this case.

The rest of the proof should be pretty straightforward. The only reason I chose $1/N<\varepsilon/6$ is because I had already completed the proof before writing this answer and knew the right choice of $N$ to pick so that I ended up with $(n+5)/(n^3-5)<\varepsilon$. However, we could have had $(n+5)/(n^3-5)<c\varepsilon$ where $c$ is any positive constant and the proof would still be acceptable.

I tried to make the actual proof as simple as possible, by attempting to minimize the amount of inequalities we had to verify, so hopefully this helps.