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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.

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  • $\begingroup$ Taking $\delta=\min{\left\{\frac{\varepsilon}2,\frac14\right\}}$ instead of just $\frac{\varepsilon}2$ would do $\endgroup$ – jdoicj Feb 13 '14 at 15:29
  • $\begingroup$ Why exactly is it a quarter? That means for the interval $(-\frac34, -\frac14)$, $\delta=\frac14$ is the required result? $\endgroup$ – David Feb 13 '14 at 15:34
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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.

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  • $\begingroup$ So then the interval $(-\infty, -\frac14)$ is not relevant for our proof? $\endgroup$ – David Feb 13 '14 at 16:05
  • $\begingroup$ @David. It's not relevant because of our particular choice of $\delta$. $\endgroup$ – jdoicj Feb 25 '14 at 5:20
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You could have just done this: $$\lim_{x\to0} \frac{(x-1)(x+1)}{(2x+1)(x-1)}$$ $$\lim_{x\to0} \frac{(x+1)}{(2x+1)}$$ $$\frac{0+1}{0+1}=1$$

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    $\begingroup$ Yes, I know but we have to prove it using the definition of a limit. Should've made that clearer in my post. $\endgroup$ – David Feb 13 '14 at 15:23
  • $\begingroup$ After cancelling out $(x-1)$ do the following.Consider two cases. Substitute $h^+$ in one case find limit which you will get as $1$. Then substitute $h^-$ and find the limit which you will get as $1$ again. $\endgroup$ – lsp Feb 13 '14 at 15:34

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