0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I think there is a factor of 4 missing in the numerator. (Also, given an $\epsilon>0$, you want to find a corresponding $\delta$.) $\endgroup$ – user84413 Apr 6 '15 at 16:49
  • $\begingroup$ There is certainly not a factor of 4 missing in the numerator. Do the algebra out. $\endgroup$ – Paddling Ghost Apr 6 '15 at 16:51
  • $\begingroup$ @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? $\endgroup$ – Rousseau Apr 6 '15 at 16:55
  • $\begingroup$ 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.) $\endgroup$ – Jonas Meyer Apr 7 '15 at 1:46
  • $\begingroup$ 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 $\epsilon$, $\endgroup$ – hardmath Apr 7 '15 at 1:48
1
$\begingroup$

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 $\epsilon$.

$\endgroup$
  • $\begingroup$ Shouldn't it be max of $\left \{2,1+\frac{3}{4(\epsilon -1)}\right\}$? And why does N have to be an integer? $\endgroup$ – Rousseau Apr 6 '15 at 17:14
  • $\begingroup$ 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. $\endgroup$ – kobe Apr 6 '15 at 17:18
  • $\begingroup$ 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. $\endgroup$ – Paddling Ghost Apr 6 '15 at 17:20
  • $\begingroup$ But doing it my way, would that mean that if I choose $\delta = max(2, \frac{1}{\epsilon})$ then $x > \delta \rightarrow \frac{1}{x} < \epsilon$ ? $\endgroup$ – Rousseau Apr 6 '15 at 17:28
  • $\begingroup$ @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. $\endgroup$ – Paddling Ghost Apr 6 '15 at 17:29
1
$\begingroup$

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

$\endgroup$

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.