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Verify: $$\lim_{x\rightarrow0}(4x^2+2x+5)=5$$ On a simple linear function it's easy to use the limit definition "$|f(x)-L|$ becomes arbitrarily small" but it won't work in this situation. But I'm supposed to use that definition somehow. The function won't factor. I'm not sure what to do.

We haven't learned limit laws yet so my professor wouldn't accept that.

EDIT: Changed my limit definition. Got it mixed up.

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    $\begingroup$ Wait, what's to stop you from just plugging the zero in? The limit isn't of indeterminate form... $\endgroup$ – recursive recursion Aug 23 '14 at 23:24
  • $\begingroup$ That wouldn't be a "proof" $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 23 '14 at 23:26
  • $\begingroup$ @Zack: Check this. $\endgroup$ – Mhenni Benghorbal Aug 23 '14 at 23:32
  • $\begingroup$ We haven't gotten to delta-epsilon proofs yet $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 23 '14 at 23:40
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    $\begingroup$ @Zack If you haven't gotten to $\epsilon-\delta$ proofs yet, how do you define limits? You can't possibly expect to prove a statement involving limits if you haven't precisely defined what a limit is! If you give us your definition of a limit, that would make answering your question easier for us to write and more understandable for you. $\endgroup$ – beep-boop Aug 23 '14 at 23:50
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Given any $\epsilon > 0$, let $\delta = \min\{1, \epsilon/6\} > 0$. Then observe that if $0 < |x - 0| < \delta$, then: \begin{align*} |(4x^2 + 2x + 5) - (5)| &= |4x^2 + 2x| \\ &= |x(4x + 2)| \\ &= |x||4x + 2| \\ &< \frac{\epsilon}{6}|4x + 2| &\text{since } |x - 0| < \delta \leq \frac{\epsilon}{6} \\ &\leq \frac{\epsilon}{6}(|4x| + |2|) &\text{by the triangle inequality} \\ &= \frac{\epsilon}{6}(4|x| + 2) \\ &< \frac{\epsilon}{6}(4(1) + 2) &\text{since } |x - 0| < \delta \leq 1 \\ &= \frac{\epsilon}{6}(6) \\ &= \epsilon \end{align*} as desired.

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  • $\begingroup$ ok I think $|x||4x+2|$ is the answer my professor is looking for $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 24 '14 at 0:16
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I think a slightly simpler choice than the other answers is $\delta = \min\{\sqrt{\epsilon/8}, \epsilon/4\}$. Then if $|x - 0| = |x| < \delta$,

$$ |(4x^2 + 2x + 5) - 5| = |4x^2 + 2x| \le |4x^2| + |2x| < \epsilon/2 + \epsilon/2 = \epsilon. $$

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    $\begingroup$ Should that be $\sqrt{\epsilon / 8}$ instead of $\sqrt\epsilon / 8$? $\endgroup$ – Théophile Aug 23 '14 at 23:56
  • $\begingroup$ @Théophile Indeed, thanks! $\endgroup$ – Sasho Nikolov Aug 24 '14 at 5:38
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Have you shown in class that polynomials are continuous everywhere? If not, see this post for a sketch of a proof by induction on the polynomial degree of this extremely useful fact.

Next, if a function $f$ is continuous at $c$, then $$\lim_{x\to c} f(x) = f(c).$$ This is exactly the definition of continuity. So you can just plug in the 0 into your polynomial.

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  • $\begingroup$ Actually that is in the next section of our book, on limit laws. We are not supposed to use those. But I see that that is an acceptable answer otherwise $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 23 '14 at 23:37
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    $\begingroup$ Adriano has posted a direct proof above. One lesson to take away from this exercise is that proving general theorems, while sometimes more work up front, will then save you a lot of trouble later when you need to compute special cases! $\endgroup$ – user7530 Aug 23 '14 at 23:42

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