Epsilon-n proof confusion

Question

So, I'm told I did something wrong on this epsilon-n proof but I really don't know what.

I'm told to prove that $$\lim_{n\to\infty} \frac{5n+1}{n+3} = 5 $$

I begin by setting $\epsilon > 0$ and searching for $k$ such that for every $ n > k$, $$\left|\frac{5n+1}{n+3} - 5\right| < \epsilon$$ $$\left|\frac{5n+1}{n+3} - 5\right| = \left|\frac{5n+1 - (5n+15)}{n+3}\right| = \left|\frac{-14}{n+3}\right| = \frac{14}{n+3} < \epsilon$$

(the last equality being valid since $n>0$ is definitely true)

$$14 < n\epsilon + 3\epsilon$$ $$ n > \frac{14-3\epsilon}{\epsilon}$$ So set $k = \left \lceil{\frac{14-3\epsilon}{\epsilon}}\right \rceil$ and for every n > k, $$\left|\frac{5n+1}{n+3} - 5\right| < \epsilon$$ by what's shown above.

The issue I'm having is that I'm told (by my instructor, who deducted a point from this problem because of it) a better $k$ is in fact $\left \lceil{\frac{14}{\epsilon}}\right \rceil$, and I don't know why that would be or how I would get that $k$ to begin with.

Answer 1

The problem is not that the lower bound is "wrong"; it is that the lower bound is unnecessarily complicated.

For all $n \geq 1$ we have $$ \bigg| \frac{5n+1}{n+3} - 5 \bigg| = \bigg|\frac{5n+1 - 5n - 15}{n+3} \bigg| = \frac{14}{n+3} < \frac{14}{n}; $$ from this it is clear that if $\varepsilon > 0$, then to make $14/n < \varepsilon$ it suffices to make $n > 14/\varepsilon$; hence for every $\varepsilon > 0$, we have $n \geq \lceil 14/\varepsilon \rceil + 1$ only if $$ \bigg| \frac{5n+1}{n+3} - 5 \bigg| < \varepsilon. $$

Answer 2

Hint: Use the inequality $$\frac{14}{n + 3} < \frac{14}{n} < \epsilon.$$ to determine $k$.

Answer 3

Contrary to what has been said in the comments on StatGod's answer, I would say the instructor's value for $k$ is better than yours. It's equally good for the proof and it makes the working simpler; how is that not better?

However I would also say that your instructor was being extremely harsh in deducting marks for this.

On the other hand, it's worth noting that in other cases, the "straightforward" approach you have given may not quite work. For example, show that $$\lim_{n\to\infty}\frac{n^2}{n^2+1}=1\ .$$ We have $$\left|\frac{n^2}{n^2+1}-1\right|=\frac1{n^2+1}\ ,$$ and making this less than $\epsilon$ in the "obvious" way gives $$n>\sqrt{\frac1\epsilon-1}\ .$$ Problem is, the RHS is meaningless if $\epsilon>1$. Now in a sense this is irrelevant since we are interested in small $\epsilon$, but on the other hand if we want to do things precisely we must recognise that the property is supposed to be true for all positive $\epsilon$. So in this case it is definitely better to say that $$\frac1{n^2+1}<\epsilon$$ is true if $$\frac1{n^2}<\epsilon$$ which is true if $$n>\frac1{\sqrt\epsilon}\ .$$