Prove, using the the $(\varepsilon-\delta)$ definition of limit that:
$$\lim_{x\to0} \frac{x^2-1}{2x^2-x-1}=1$$
What I've tried so far:
By the $\varepsilon-\delta$ definition of a limit $$0<|x|<\delta\implies\bigg|\frac{x^2-1}{2x^2-x-1}-1 \bigg|<\varepsilon$$ so I'm trying to express the RHS in terms of $|x|$: $$\bigg|\frac{x^2-1}{2x^2-x-1}-1\bigg|\iff\bigg|\frac{-x^2+x}{2x^2-x-1}\bigg|\iff\frac{|x||x-1|}{|2x+1||x-1|}\iff\frac{|x|}{|2x+1|}<\frac{\delta}{|2x+1|}$$ I know I should somehow get rid of the variable $x$ by making an estimation on $|2x+1|$. If I estimate that: $|2x+1|>\frac12$ which holds for $x\in(-\infty,-\frac34)\cup(-\frac14,+\infty)$ I get: $$\frac{\delta}{|2x+1|}<\frac{\delta}{\frac12}=2\delta$$So then $\delta=\frac\varepsilon2$.
Is the esimation justified? Should I also find a suitable $\varepsilon$ for the interval $(-\frac34, -\frac14)$? Or maybe there is anoher method on proving the statement.
I'd appreciate some suggestions on how to proceed or correct my proof.
We have to show that a $\delta$ exists for every $\varepsilon$ so that for every $x$ belonging to the interval $-\delta<x<\delta$ satisfies $|f(x)-l|<\varepsilon$.
You started with an inequality that holds in the region $\left(-\frac14,\infty\right)$ and proved at any $\delta \le \frac\varepsilon2$ would make $|f(x)-l|<\varepsilon$. So all you need is a $\delta$ so that $x>-\frac14$ and $\delta \le \frac\varepsilon2$
And now, $\delta=\min\{\frac{1}{4},\frac\varepsilon2\}$ is one such delta.