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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

You could have just done this: $$\lim_{x\to0} \frac{(x-1)(x+1)}{(2x+1)(x-1)}$$ $$\lim_{x\to0} \frac{(x+1)}{(2x+1)}$$ $$\frac{0+1}{0+1}=1$$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.