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How can I solve this using the definition of limit?

Prove using the definition of limit that: $$\lim_{x\to 1} (x²-4x)=-3$$

How can I approach this?

EDIT: OH my god! Thanks @adam!

Maybe you can also help me on out on that one:

$$\lim_{x\to -\infty} \frac{1}{x+2}=0$$

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  • $\begingroup$ I would start by following the worked example here: en.wikipedia.org/wiki/… $\endgroup$ – Mark B Jul 9 '14 at 23:34
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    $\begingroup$ Generally better to ask a separate question rather than wait for an answer to the first and then tack on another question. $\endgroup$ – Thomas Andrews Jul 10 '14 at 0:03
  • $\begingroup$ Chech this. $\endgroup$ – Mhenni Benghorbal Jul 10 '14 at 0:22
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Let $\epsilon >0$ be given. Then consider the quantity

$$|x^2-4x+3|=|x-3||x-1|$$

If we know that $|x-1|<\delta$, then we see this means

$$|x-3| = |x-1-2|$$ $$\le |x-1|+2$$ $$<2+\delta.$$

Hence we see

$$|x^2-4x+3|<\delta^2+2\delta$$

so if we do

$$(\delta+1)^2=\delta^2+2\delta+1=\epsilon+1$$

we see this gives a positive value

$$\delta=-1+\sqrt{1+\epsilon}.$$

we have the right result.

For the second one it's even easier. Let $\epsilon >0$ be given.

Then we see that $$\left|{1\over x+2}\right|<\epsilon\iff |x+2|>{1\over\epsilon}$$

Since $x\to -\infty$, we can assume $x<0$ so that $|x+2|=|x|-2$, so choose $N={1\over\epsilon}+2$ and for $x< N$ we have the desired result.

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  • $\begingroup$ Thanks! @adam! I edited the question, maybe you can help me out on this other one too, the minus infinity is dificulting the calculus. $\endgroup$ – user136829 Jul 9 '14 at 23:53
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    $\begingroup$ @user136829 see my edited answer. Also, after this it's probably best to accept an answer and write a new question if you have more to do, it helps avoid constant edits. $\endgroup$ – Adam Hughes Jul 10 '14 at 0:01
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Alternative to Adam's answer: First, ensure that $\delta<1$. So if $|x-1|<1$ then $-1<x-1<1$ so $-3<x-3<0$ and hence $|x-3|<3$.

Now use Adam's approach: $\left|x^2-4x-(-3)\right|=|x-1|\,|x-3|<3\delta$. So you can pick $\delta=\min(1,\epsilon/3)$.

That $\min$ trick is pretty common in limits and worth knowing.

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  • $\begingroup$ Yeah, op: this is worth mentioning since it's not always possible to solve for the optimal $\delta$ like I did there. $\endgroup$ – Adam Hughes Jul 10 '14 at 0:08
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A more intuitive way to proceed: Let $\epsilon > 0$, we are looking for $\delta > 0$ such that if $|x-1|\leq \delta$, then $|x^2-4x-(-3)|< \epsilon$. Note that

$$|x^2-4x+3| = |x-1-2||x-1| \leq (|x-1|+2)|x-1| <(\delta + 2)\delta,$$

so if $\delta = \min\{1,\frac{\epsilon}{3}\}$ then $$(\underbrace{\delta}_{\leq 1} + 2)\underbrace{\delta}_{\epsilon/3} \leq 3\frac{\epsilon}{3}= \epsilon.$$ And the proof is done.

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You need to find $\delta > 0$ such that if $|x - 1| < \epsilon$, then $|x^2 - 4x -(-3)| < \epsilon$. The idea is to keep bounding $|x^2 - 4x -(-3)|$ until you get something that just is $\delta$ involving some constants whatsoever. Notice that $x-1$ is a factor of $x^2 - 4x + 3$, since the limit is true. I'd rather not do everything for you, since this kind of exercise is very important to get used to manipulating epsilons and deltas. But I answered a question, giving the general strategy to deal with limits of polynomials. I believe that if you read it carefully, you can solve it yourself. If you have difficulties, please say, and I'll elaborate more on your specific problem.

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