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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)[reply]

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)[reply]

\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)[reply]

Well, we have

x

in the interval

(1-\delta ,1+\delta ),

where

δ

is yet to be determined. So

5 x - 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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

@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) [reply]