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This is a very simple question. I want to prove that $\frac{4n+3}{7n-10} \to \frac{4}{7}$

I did the algebra, that is $\left | \frac{4n+3}{7n-10} - \frac{4}{7} \right | = \left |\frac{61}{49(n-\frac{10}{7})} \right | < \epsilon$

Now the answer key did something very interesting. They underestimated the fraction to get a neater upper bound. They did $\frac{61}{49(n-\frac{10}{7})} \leq \frac{61}{n} < \epsilon$.

Then they chose $N = \max\left \{ 10,61/\epsilon\right \}$.

I was lazy so I decided to not underestimated the fraction. Instead, I just took $n > N = \frac{61}{49\epsilon}+\frac{10}{7}$. Is my N big enough for the proof to work?

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I think you can answer this question yourself. If $n > \frac{61}{49\epsilon} + \frac{10}{7}$, then does it follow that $|\frac{61}{49(n - \frac{10}{7})}|<\epsilon$? –  Neal Oct 13 '12 at 1:26
Wait, isn't estimating the fraction the lazy thing to do, since it involves doing less work with simpler expressions? :p P.S. I'm skeptical that your $N$ works when $\epsilon$ is very large, since you ignore the effect of the absolute value signs. –  Hurkyl Oct 13 '12 at 1:27
It does, it works "too perfectly". –  Hawk Oct 13 '12 at 1:27

2 Answers 2

up vote 1 down vote accepted

$$n > \frac{61}{49\epsilon}+\frac{10}{7} \implies \frac{49\left(n-\frac{10}{7}\right)}{61} > \frac{1}{\epsilon}$$ Assuming $n > \frac{10}{7}$, you have the desired inequality, so it is correct. Although simplifying it via underestimation certainly is a lot cleaner.

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It Should work. You have to consider the integer part of this choice of N($\epsilon$).

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