Could someone verify this proof? Prove $\lim\limits_{n\to\infty} C_n = \lim\limits_{n\to\infty} \dfrac{4n+3}{7n-5} = \dfrac{4}{7}$
Proof: Let $\epsilon > 0$ and take $N = \dfrac{41}{49\epsilon} + 1$ This implies that for all $n > \dfrac{41}{49\epsilon} + 1 \iff n-1 > \dfrac{41}{49\epsilon} \iff \epsilon(n-1) > \dfrac{41}{49} \iff \epsilon > \dfrac{41}{49n - 49} \iff$
$ \iff \epsilon > \dfrac{41}{49n-35} \iff \epsilon > \dfrac{28n+21-28n+20}{49n-35} \iff \dfrac{7(4n+3)-4(7n-5)}{7(7n-5)} \iff \epsilon > \dfrac{4n+3}{7n-5}-\dfrac{4}{7} \iff \epsilon >|\dfrac{4n+3}{7n-5}-\dfrac{4}{7}|$
End of proof