Find $\lim_{n \to \infty} \frac{5n^{3} + 3n - 7}{11n^{3}+n}$ and use the definition of the limit to verify that your answer is correct.
I found that the limit = $\frac{5}{11}$ and verified it as follows: $$ \biggl| \frac{5n^{3} + 3n - 7}{11n^{3}+n} - \frac{5}{11} \biggr| = \frac{28n-77}{121n^3+11n} < \frac{28n}{121n^3 +11n} = \frac{28}{121n^2 + 11n} < \frac{28}{121n^2} < \frac {28}{84n^2} = \frac{1}{3n^2} = \epsilon$$
Then $\forall \epsilon, n = (3\epsilon)^{1/2} $. Is this a valid verification? Do I need to go any further to justify that this is the limit?
I also do not quite understand why I can continue to produce a chain of less than inequalities and still have the statement be true for any $\epsilon$ ?
In general, $\sqrt{3\varepsilon}\notin\mathbb N$. So, you have a problem here.
Note that the inequality$$\left\lvert\frac{5n^3+3n-7}{11n^3+n}-\frac5{11}\right\rvert=\frac{28n-77}{121n^3+11n}$$is trivially false if $n=1$ or $n=2$. Otherwise, it is correct.
You should end the proof with: take $N\in\mathbb N$ such that $\frac1{3N^2}<\varepsilon$. Then$$n\geqslant N\implies\frac1{3n^2}\leqslant\frac1{3N^2}<\varepsilon.$$