This was a question on my Real Analysis exam that I initially did wrong, but the instructor provided no feedback except "?" so I want to make sure I did it correctly this time. (For clarity, this is not what I put on the exam - I initially found a value for $N$ that didn't work, so this is my correction)
Use $\varepsilon$-$N$ definition of convergence to show $\lim_{n \to \infty} (1+\frac{1}{n})^2 = 1$
Side work: $$(1 + \frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2}$$
$$\left|(1 + \frac{1}{n})^2 - 1 \right| = \left|1 + \frac{2}{n} + \frac{1}{n^2} - 1 \right| = \frac{2}{n} + \frac{1}{n^2} < \frac{2}{n} + \frac{1}{n} = \frac{3}{n}$$
$$\frac{3}{n} < \varepsilon \implies n > \frac{3}{\varepsilon} \approx N$$
Proof: Given $\varepsilon > 0$, set $N = \left[ \frac{3}{\varepsilon} \right] +1$. Then $\forall$ $n>N$ we have:
$$\left|(1 + \frac{1}{n})^2 - 1 \right| = \left|1 + \frac{2}{n} + \frac{1}{n^2} - 1 \right| = \frac{2}{n} + \frac{1}{n^2} < \frac{2}{n} + \frac{1}{n} = \frac{3}{n} < \frac{3}{N} = \frac{3}{\left[ \frac{3}{\varepsilon} \right] +1} < \frac{3}{\frac{3}{\varepsilon}} = \varepsilon$$
Conclude by the $\varepsilon$-$N$ definition of convergence that $\lim_{n \to \infty} (1 + \frac{1}{n})^2 = 1$. $\Box$
Is this correct?