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 ?


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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $\delta$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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