# Proving limit with epsilon delta

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 6 '15 at 16:49
• There is certainly not a factor of 4 missing in the numerator. Do the algebra out. – Paddling Ghost Apr 6 '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 6 '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 7 '15 at 1:46
• 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 7 '15 at 1:48

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 $\epsilon$.

• 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 6 '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 6 '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 6 '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 6 '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 6 '15 at 17:29

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