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

= February 26 =

Limit of quotient
How to prove $$\lim_{x\to1}\frac{5x^2+5}{5x-4}=10$$ 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 $&epsilon; &gt; 0$ ..." – the definition of a limit is going to require that you find a good enough $&delta;$ for a given, arbitrary $&epsilon;$.  No deductions (or assumptions, if that's what you meant) about the value of $&epsilon;$ 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 &bull; 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 $&delta;$ is yet to be determined. So $5x &minus; 4$ 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 $&delta;$ 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 $&delta;$ 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 $&delta;$ to be less than 1/10, then the denominator will always be at least 1/2, etc.  –Deacon Vorbis (carbon &bull; 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
 * $$\lim_{x\to1}\frac{5x^2+5}{5x-4}=1+\lim_{x\to1}\frac{9}{5x-4}$$
 * So, you only need to prove that
 * $$\lim_{x\to1}\frac{1}{5x-4}=1$$,
 * 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)
 * 1. Please ping the user to whom you answer. 2. You've put the closing tags twice and no opening tag. --CiaPan (talk) 11:30, 27 February 2019 (UTC)