Find $N \gt 0$ so that $ n \ge N \implies \lvert \left(1+\frac3n \right)^2 -1\rvert \lt \frac{1}{10}$. Here's the problem:
Find $N \gt 0$ so that
$$ n \ge N \implies \left\lvert \left(1+\frac3n \right)^2 -1\right\rvert \lt \frac{1}{10}$$
So, I'm not quite sure what this is asking me. I found a value of $n$ (e.g. $100$ works) that makes the consequent true, but I don't think that simply choosing any value for $N$ less than or equal to $100$ to make the antecedent true solves the problem. As in I don't see how the antecedent would imply the consequent in that case; it'd just be vacuously true if both propositions were true. Or am I going about this all wrong? Any help would be appreciated. Thanks.
 A: Let $m = 1+\dfrac{3}{n} \to |m^2-1| < \dfrac{1}{10} \to -\dfrac{1}{10} < m^2-1 < \dfrac{1}{10} \to \dfrac{9}{10} < m^2 < \dfrac{11}{10} \to \dfrac{3}{\sqrt{10}} < m < \dfrac{\sqrt{11}}{\sqrt{10}}$. Observe that $\forall n \geq 1 \to m = 1+\dfrac{3}{n} > \dfrac{3}{\sqrt{10}}$. And $m < \dfrac{\sqrt{11}}{\sqrt{10}} \to \dfrac{3}{n} < \dfrac{\sqrt{11}-\sqrt{10}}{\sqrt{10}}\to n > \dfrac{3\sqrt{10}}{\sqrt{11}-\sqrt{10}}=3\sqrt{110}+30\approx 61.5$. Thus $N = 62$ or higher.
A: $$\begin{align*}\left\lvert \left(1+\frac3n \right)^2 -1\right\rvert \lt \frac{1}{10} &\iff -\frac{1}{10} \lt \left(1+\frac3n \right)^2-1 \lt \frac{1}{10}\\& \iff \frac{9}{10} \lt \left(1+\frac3n \right)^2 \lt \frac{11}{10} \\& \iff \sqrt{\frac{9}{10}} \lt 1+\frac3n \lt \sqrt{\frac{11}{10}} \\& \iff \sqrt{\frac{9}{10}}-1 \lt \frac3n \lt \sqrt{\frac{11}{10}}-1 \\& \implies  \frac n3 \gt \frac{1}{\sqrt{\frac{11}{10}}-1} \\&\implies n \gt \frac{3}{\sqrt{\frac{11}{10}}-1}\end{align*}$$ so $n$ has to be at least $$n \ge \left \lceil \frac{3}{\sqrt{\frac{11}{10}}-1} \right \rceil= \lceil 61.46 \rceil =62$$
A: You can set the following equal to each other, that is,
$$\left\lvert \left(1+\frac3N \right)^2 -1\right\rvert = \frac{1}{10}$$
You will find that $N \approx 61.4643$.
