Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

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

share|cite|improve this question

closed as off-topic by T. Bongers, alexqwx, This is much healthier., M Turgeon, Claude Leibovici Jul 10 '14 at 4:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, alexqwx, Community, M Turgeon, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

I would start by following the worked example here:… – Mark B Jul 9 '14 at 23:34
Generally better to ask a separate question rather than wait for an answer to the first and then tack on another question. – Thomas Andrews Jul 10 '14 at 0:03
Chech this. – Mhenni Benghorbal Jul 10 '14 at 0:22
up vote 4 down vote accepted

Let $\epsilon >0$ be given. Then consider the quantity


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


so if we do


we see this gives a positive value


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.

share|cite|improve this answer
Thanks! @adam! I edited the question, maybe you can help me out on this other one too, the minus infinity is dificulting the calculus. – user136829 Jul 9 '14 at 23:53
@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. – Adam Hughes Jul 10 '14 at 0:01

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.

share|cite|improve this answer
+1 because you got the same answer as me :) – Surb Jul 10 '14 at 0:07
Yeah, op: this is worth mentioning since it's not always possible to solve for the optimal $\delta$ like I did there. – Adam Hughes Jul 10 '14 at 0:08

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.