Wikipedia:Reference desk/Archives/Mathematics/2019 February 26

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February 26

Limit of quotient

How to prove $\lim _{x\to 1}{\frac {5x^{2}+5}{5x-4}}=10$ by $\varepsilon ,\delta$?

I like proofs which avoid using arbitrary deductions such as $\varepsilon =1$ etc. יהודה שמחה ולדמן (talk) 22:29, 26 February 2019 (UTC)

What do you mean by "arbitrary deductions"? Any valid proof is going to start off with something like "Let ε > 0 ..." – the definition of a limit is going to require that you find a good enough δ for a given, arbitrary ε. No deductions (or assumptions, if that's what you meant) about the value of ε can be made, except that it's positive. Also, rather than someone here just doing the proof for you, it's generally good if you can say a little about what you've tried, what doesn't work, where you're stuck, etc. –Deacon Vorbis (carbon • videos) 23:26, 26 February 2019 (UTC)

$\left|{\frac {5x^{2}+5}{5x-4}}-10\right|={\frac {5|x-1||x-9|}{|5x-4|}}$

How do I get rid of the denominator? I tried a few inequalities with no success. יהודה שמחה ולדמן (talk) 00:04, 27 February 2019 (UTC)

Well, we have x in the interval $(1-\delta ,1+\delta ),$ where δ is yet to be determined. So 5x − 4 must be somewhere in the interval $(1-5\delta ,1+5\delta ).$ We need to bound it away from 0, so we just need to pick a value for δ which will keep the the lower end of that interval positive – anything (strictly) less than 1/5 will work. For example, if you require your δ to always be less than 1/6, then the denominator will always be at least 1/6 (and 1 over the denominator less than 6); if you require δ to be less than 1/10, then the denominator will always be at least 1/2, etc. –Deacon Vorbis (carbon • videos) 01:50, 27 February 2019 (UTC)

To be pedantic, for any dense set A (such as the rational numbers), we can assume without loss of generality that $\epsilon \in A$.--Jasper Deng (talk) 03:16, 1 March 2019 (UTC)

This is rather a contrived example. You can note that

$\lim _{x\to 1}{\frac {5x^{2}+5}{5x-4}}=1+\lim _{x\to 1}{\frac {9}{5x-4}}$

So, you only need to prove that

$\lim _{x\to 1}{\frac {1}{5x-4}}=1$,

which is much simpler. Ruslik_Zero 10:16, 27 February 2019 (UTC)

The simplification relies on the the limit law about the linearity of taking the limit, which I believe is out of the scope of the OP's question, even if it makes things harder for them.--Jasper Deng (talk) 10:34, 27 February 2019 (UTC)

Also, I think your simplification assumes the numerator is $5x+5$ but it's $5x^{2}+5$, meaning it takes more than just that limit law to simplify that way.--Jasper Deng (talk) 10:48, 27 February 2019 (UTC)

@Jasper Deng: 1. Please ping the user to whom you answer. 2. You've put the closing </math> tags twice and no opening tag. --CiaPan (talk) 11:30, 27 February 2019 (UTC)