Prove. $\frac{x}{x^2-2} \rightarrow 1$ as $x \rightarrow 2$ from first principles.

Question

I'm a student studying math at university. I'm practising proving limits using first principles like in the question above and I was hoping to get some feedback on my method. I'm still getting the hang of things so my proof will likely be quite messy especially in the latter half, of which I have little experience. Any pokes or prods in the right direction would be much appreciated!

First I fix $\epsilon >0$ and $\delta > 0$ such that.

$$0<|x-2|< \delta \implies \bigg| \frac{x}{x^2-2} - 1 \bigg| < \epsilon$$

$$\implies \bigg|\frac{(x+1)(x-2)}{x^2 - 2}\bigg| < \epsilon$$

$$\implies \bigg|\frac{(x+1)}{x^2 - 2}\bigg||x-2| \leq |x+1||x-2| < \epsilon$$

Let $|x-2| < \frac{1}{2}$

$-\frac{1}{2} < x-2 < \frac{1}{2}$ or $2\frac{1}{2} < x+1 < 3\frac{1}{2} $

Let $\epsilon >0$ be given.

Choose $ \delta = \min(\frac{1}{2},\frac{2\epsilon}{7})$

Then $|x-2| < \delta \implies |x+1||x-2|< \frac{7}{2}\delta \leq \epsilon$

Thanks for your time!

Answer 1

Good work, although it is not immediately clear why, $$\left| \frac{(x+1)}{x^{2}-2} \right| |x-2|\leq |x+1||x-2|$$ When $x \geq \sqrt{3}$, this is certainly true. But I do feel that this merits more explanation.