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?