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We are given that $g(x) < f(x) < h(x)$ over some interval, $$\lim _{x\rightarrow a} g(x)=L$$ and $$\lim _{x\rightarrow a} h(x)=L$$

Through this we can infer that for every $ε_h$ there exists a $δ_h$ such that $|h(x)-L|<ε_h$ when $0<|x−a|<δ_h$ and that for every $ε_g$ there exists a $δ_g$ such that $|g(x)-L|<ε_g$ when $0<|x−a|<δ_g$

Let $δ$ be defined as $\min(δ_h,δ_g)$

We can subtract $L$ from each term of the given inequality to get $g(x)-L < f(x)-L < h(x)-L$

We get $-ε_g<g(x)-L < f(x)-L < h(x)-L<ε_h$. Now we define $ε$ as $\max(ε_h,ε_g)$. Using this we get $|f(x)−L|<ε$ if $0<|x−a|<δ$

This Completes the proof.

I wanted to ask 1) whether the proof is correct? 2) whether it's easy to follow ?

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While you get good ideas, it is not easy to follow your proof.

In general, when you want to prove convergence using the $\epsilon - \delta$ rule, you have to fix $\epsilon >0$ and to find $\delta$.

So here you should:

  • Select $\epsilon > 0$.
  • For this $\epsilon$, you'll be able to find $\delta_g$ such that $|g(x)-L|<ε$ for $0<|x−a|<δ_g$. Similarly, you'll find $\delta_h$.
  • Now for $\delta = \min (\delta_g, \delta_h)$, you have $-ε<g(x)-L < f(x)-L < h(x)-L<ε$ as you noticed providing that $0<|x−a|<δ$.
  • This concludes the proof.

The introduction of $\epsilon_g, \epsilon_h$ is not necessary. $\epsilon$ is sufficient.

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  • $\begingroup$ Alright I can see why introducing more epsilons is confusing. Is my proof correct though? $\endgroup$ – Star Platinum ZA WARUDO Aug 19 '18 at 15:52
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    $\begingroup$ I would say no. Because you can't define $\epsilon$ like you did at the end. You have to pick it up at the start and from it find an appropriate $\delta$. $\endgroup$ – mathcounterexamples.net Aug 19 '18 at 15:53
  • $\begingroup$ Think of it like this You give me a positive number ($ϵ$) I assign that as $ϵ_g$ and assign a random positive number to $ϵ_h$. Then I find a delta for both of them and give you whichever is the lesser one. Since I can do this with all positive numbers, the limit is correct. $\endgroup$ – Star Platinum ZA WARUDO Aug 19 '18 at 16:10
  • $\begingroup$ Why the downvotes?? This is genuine and correct answer. +1 to compensate. $\endgroup$ – Paramanand Singh Aug 19 '18 at 16:16
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The proof is not correct. Your $\epsilon$ depends on the other two $\epsilon$ which makes the proof questionable

It is readable and it shows your good writing skills.

You may improve your proof by starting with an arbitrary $\epsilon$ and find your $\delta$

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