Evaluate the $\lim_{n \rightarrow \infty} \frac{2n^{2} + 3}{n^{2} - 7}$ using the $\epsilon$ - definition, assuming the Archemedian property

Question

Evaluate: $$\lim_{n \rightarrow \infty} \frac{2n^{2} + 3}{n^{2} - 7}$$ using the $\epsilon$ definition, assuming the archemedian property.

I constantly keep on having problems trying to solve for an explicit N.

So by definition of a convergent sequence I am suppose to show: $$\forall \epsilon > 0, \ \exists \ N\in \mathbb{N} \ s.t. \ N \leq n \Rightarrow \Bigg|\frac{2n^{2} + 3}{n^{2} - 7} - 1\Bigg| < \epsilon$$

SO after some manipulation I arrived at the expression: $$\frac{10 + 7\epsilon}{\epsilon - 1} < N^{2}$$

So proceeding from here I would find N, but I am having issue with accepting this because if $\epsilon$ can be any arbitrary number I would be screwed if $\epsilon$ was chosen to be an value $s$ such that $s \leq 1$. Any of those values would mess up my expression I found. So what is it I am doing wrong? I didn't use the Archemedian property in my solution and don't see how it could be of use here.

Answer 1

Note that

$$\lim_{n \rightarrow \infty} \frac{2n^{2} + 3}{n^{2} - 7}=2$$

therefore we need to show

$$ \left|\frac{2n^{2} + 3}{n^{2} - 7} - \color{red}2\right| < \epsilon$$

Note also that we have

$$\frac{2n^{2} + 3}{n^{2} - 7}=\frac{2n^{2} -14+ 17}{n^{2} - 7}=2+\frac{17}{n^{2} - 7}$$

therefore it suffices show that

$$\lim_{n \rightarrow \infty} \frac{1}{n^{2} - 7}=0$$