My workings is[sic] attached below
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Prove that \(\displaystyle \displaystyle{\lim_{n\, \rightarrow \,\infty}}\, \dfrac{3n\, +\, 1}{7n\, -\, 4}\, =\, \dfrac{3}{7}\)
From the definition of a convergent sequence. Let \(\displaystyle \epsilon\, >\, 0\), we need to produce \(\displaystyle n(e)\, >\, 0\) such that:
. . . . .\(\displaystyle \left|\dfrac{3n\, +\, 1}{7n\, -\, 4}\right|\, <\, \epsilon,\, \forall\, n\, >\, n(e)\)
Therefore:
. . . . .\(\displaystyle \left|\dfrac{21n\, +\, 7\, -\, 21n\, +\, 12}{7(7n\, -\, 4)}\right|\, =\, \left|\dfrac{19}{49A\, -\, 4}\right|\)
What is your \(\displaystyle n(e)\)? What do you mean by "therefore"? What is "A"?