Prove that $\lim_{x \rightarrow \infty} \frac{x^2+3}{4x^2-4x+8} = \frac{1}{4}$ using the $\epsilon$-$\delta $ definition of a limit.

Question

Prove that $$\lim_{x \rightarrow \infty} \frac{x^2+3}{4x^2-4x+8} = \frac{1}{4}$$ using the $\epsilon$-$\delta $ definition of a limit. So we want to find a $\delta$ such that for every $\epsilon>0$ $$x > \delta \rightarrow \left|\frac{x^2+3}{4x^2-4x+8} - \frac{1}{4}\right| = \left|\frac{x+1}{4x^2-4x+8}\right| < \epsilon. $$ The ratio has a global maximum at $x=2$ Thus, $x+1 < 2x$ and $4x^2-4x+8 > 2x^2 + 8 > 2x^2$. This implies that with $ \delta >2$ $$ x > \delta \rightarrow |\frac{x+1}{4x^2-4x+8}| < \frac{2x}{2x^2} = \frac{1}{x}. $$ I think this proves it... How do I wrap it up?

I think there is a factor of 4 missing in the numerator. (Also, given an $\epsilon>0$, you want to find a corresponding $\delta$.) — user84413, Apr 06 '15 at 16:49
There is certainly not a factor of 4 missing in the numerator. Do the algebra out. — Paddling Ghost, Apr 06 '15 at 16:51
@MagicMan Maybe realizing that the ratio assumes a global maximum for x=2? So I guess $\delta$>2 but I how do I simplify my expression so I can end up with something meaningful? — Rousseau, Apr 06 '15 at 16:55
I will be a bit more explicit: "So we want to find a $\delta$ such that for every $\varepsilon>0$..." is incorrect, in that you cannot choose $\delta$ independent of $\varepsilon$, except when the function is eventually constant. (Also it is unusual to use $\delta$ for this.) — Jonas Meyer, Apr 07 '15 at 01:46
Your word order suggests that "we want to find a $\delta$" that works "for every $\epsilon \gt 0$", but the definition of limit allows $\delta$ values that depend on the choice of $\epsilon$, — hardmath, Apr 07 '15 at 01:48
Does this answer your question? Does there exist an epsilon-delta proof of the limit for all polynomials? — , Oct 08 '23 at 21:34

kobe · Accepted Answer · 2015-04-06T17:36:17.623

2

Note that for $x > 2$,

\begin{align}\tag{*}\left|\frac{x^2 + 3}{4x^2 - 4x + 8} - \frac{1}{4}\right| &= \left|\frac{x + 1}{4x^2 - 4x + 8}\right|\\ & = \frac{x + 1}{4(x^2 - x + 2)} \\ &< \frac{x + 1}{4(x - 1)^2}\\ & = \frac{1}{4(x - 1)} + \frac{1}{2(x - 1)^2} \\ &< \frac{3}{4(x - 1)}. \end{align}

Given $\epsilon > 0$, let $M = \max\{2,1 + \frac{3}{4\epsilon}\}$. If $x > M$, then the left-hand side of $(*)$ is less than $\frac{3}{4(x-1)}$, which is less than $\frac{3}{4(M-1)}$, which is less than $\epsilon$.

edited Apr 06 '15 at 17:36

answered Apr 06 '15 at 17:07

kobe

41,901

Shouldn't it be max of $\left {2,1+\frac{3}{4(\epsilon -1)}\right}$? And why does N have to be an integer? – Rousseau Apr 06 '15 at 17:14
Note that if $x > 2$, $$\frac{3}{4(x-1)} < \epsilon \iff \frac{3}{4\epsilon} < x - 1 \iff 1 + \frac{3}{4\epsilon} < x.$$ This is why I've made $N$ greater than $\max{2, 1 + \frac{3}{4\epsilon}}$. Depending on your definition of a limit as $x\to \infty$, you either have $N$ to be a positive integer are just a positive real number. – kobe Apr 06 '15 at 17:18
certainly not, If $N > 1+ \frac{3}{4 \epsilon}$ then $\epsilon > \frac{3}{4(N-1)}$. Not sure about the integer thing here. It works if its an integer, not sure if it needs to be. – Paddling Ghost Apr 06 '15 at 17:20
But doing it my way, would that mean that if I choose $\delta = max(2, \frac{1}{\epsilon})$ then $x > \delta \rightarrow \frac{1}{x} < \epsilon$ ? – Rousseau Apr 06 '15 at 17:28
@kobe 1) i was responding to rosseau's question in the comment, not yours. 2) I say "if" , so i'm assuming that max of ${2,1+\frac{3}{4 \epsilon}}$ is $1+\frac{3}{4 \epsilon}$. 3. i fixed the "equals" to greater than, i missed that in your step. – Paddling Ghost Apr 06 '15 at 17:29
@user3440448 ok I understand. – kobe Apr 06 '15 at 17:30
@Rousseau yes, choosing $\delta = \max(2, \frac{1}{\epsilon})$ will do. And since you don't use the $\epsilon-N$ formulation of a limit, I'll make a small edit to my answer. – kobe Apr 06 '15 at 17:35
Not sure where the downvote on this answer is coming from. I don't think the OP has demonstrated that he/she is a help-vampire. They have taken steps to edit the question to include more work, and demonstrated that they are not simply copying the answer for homework. Further, the answer is well thought out and prompted a very helpful discussion of limits. – Paddling Ghost Apr 06 '15 at 17:40
Thanks for the comment @user3440448, the two downvoters didn't explain what issues they had with my answer, and there should be more upvotes on Rousseau's question for precisely the reasons you state. – kobe Apr 06 '15 at 17:48

score 1 · Answer 2 · answered Apr 06 '15 at 17:43

We observe: $$\begin{gathered} \frac{{{x^2} + 3}}{{4{x^2} - 4x + 8}} = \frac{1}{4} \cdot \frac{{{x^2} + 3}}{{{x^2} - x + 2}} \hfill \\ \hfill \\ \frac{1}{4} \cdot \frac{{{x^2} - x + 2 + x + 1}}{{{x^2} - x + 2}} = \frac{1}{4} \cdot (1 + \frac{{x + 1}}{{{x^2} - x + 2}}) = \frac{1}{4} + \frac{1}{4} \cdot \frac{{x + 1}}{{{x^2} - x + 2}} \hfill \\ \hfill \\ \frac{{{x^2} + 3}}{{4{x^2} - 4x + 8}} - \frac{1}{4} = \frac{1}{4} \cdot \frac{{x + 1}}{{(x + 1)(x - 2) + 4}} \hfill \\ \end{gathered} $$

And for $x \geqslant 3$ we get for RHS:

$$\frac{1}{4} \cdot \frac{{x + 1}}{{(x + 1)(x - 2) + 4}} < \frac{1}{4} \cdot \frac{{x + 1}}{{(x + 1)(x - 2)}} = \frac{1}{4} \cdot \frac{1}{{x - 2}}$$

It remains to show $$\mathop {\lim }\limits_{x \to \infty } \frac{1}{{x - 2}} = 0$$

Prove that $\lim_{x \rightarrow \infty} \frac{x^2+3}{4x^2-4x+8} = \frac{1}{4}$ using the $\epsilon$-$\delta $ definition of a limit.

2 Answers2