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Question:

Let $f :(a,b)\rightarrow \mathbb{R}$ be such that $\lim_{x\to c} f(x)> \alpha$, where $c\in(a,b)$ and $\alpha\in\mathbb{R}$. Prove that there exists some $\delta > 0$ such that $$f(c+h)>\alpha\quad \text{for all $0<|h|<\delta$}.$$

My approach: Consider $c+h=x$. Then, $|x-c|<\delta$. I took $\lim_{x\to c} f(x)=l$. I then considered $\epsilon$ such that $\alpha +\epsilon =l$.

Then, I took $|f(x)-l|<\epsilon$ so $$l- \epsilon <f(x)< l + \epsilon$$

Substituted $l- \epsilon = \alpha$ to get required proof.

Is my approach correct? Is there any better way to do this?

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When someone is working out a proof, there is often a period of time during which they take formulas that look like the desired result and manipulates them in order to find some other formulas that eventually help them write a proof. But nothing they do during that time is part of the proof; it merely helps them figure out a good way to take some step when they finally write the proof.

Most of what you have written is good for that before-we-start-the-proof period of time but should not be considered part of any proof of anything.

You have a proposition that says, given a set of statements, if those statements are all true then here is another statement that also must be true.

To prove such a proposition, the first thing to do is generally to write something that says you assume the first set of statements in the proposition is true. For example, you could start your proof with something like this:

Suppose $f$ is a function, $f :(a,b)\rightarrow \mathbb R,$ suppose $c\in(a,b),$ and suppose $\alpha\in\mathbb{R}$ such that $\lim_{x\to c} f(x)$ exists and $\lim_{x\to c} f(x) > \alpha.$

This gives you that $f$ is some particular function (though you know only a little about it), and $a,$ $b,$ $c,$ and $\alpha$ are some real numbers. The given statements do not tell you exactly which numbers each of those letters represents, but for the rest of the proof you can treat $a$ (for example) as some particular number that someone else knows.

You do not yet have any numbers called $h$ or $x.$ The name $h$ has not even been mentioned, and $x$ has occurred so far only inside the expression $\lim_{x\to c} f(x),$ where it is what logicians call a "bound variable": it has a meaning inside that expression but not outside the expression. (If you have ever done any programming in a language that has local variable definitions, you can think of $x$ in $\lim_{x\to c} f(x)$ as a local variable, valid within the scope where it is defined but if you try to use it anywhere outside that scope you get an error and your program doesn't run.)

Therefore this,

Consider $c+h=x$

is fine for the before-we-start-the-proof manipulations that you do on a piece of scrap paper somewhere, but it makes no sense as the first thing in your proof or even the first thing after you have written the sentence that starts with "Suppose". Neither $h$ nor $x$ means anything yet in this context.

What actually would be good to write immediately after the "Suppose" sentence would be to spell out what $\lim_{x\to c} f(x)$ means, using the $\epsilon$-$\delta$ definition. Your statement that $\lim_{x\to c} f(x)=l$ could be part of this:

Let $\lim_{x\to c} f(x)=l.$ Then for any $\epsilon > 0,$ there exists a $\delta > 0$ such that for all $x$ such that $0 < \lvert x - c\rvert < \delta,$ we have $\lvert f(x) - l\rvert < \epsilon.$

Now comes the clever part of the proof, which all your before-the-proof manipulations helped you to figure out. If you set the value of $\epsilon$ such that $\alpha + \epsilon = l,$ you can plug that value of $\epsilon$ into the definition you just wrote, and good things will happen.

So now you can write something like

Consider $\epsilon$ such that $\alpha + \epsilon = l$

or even more simply,

Let $\epsilon = l - \alpha.$

You now have to show that $\epsilon > 0$ (an important fact that you did not write, though you probably thought about it). Once you have done that, the definition of $\lim_{x\to c} f(x)=l$ tells you there exists a $\delta$ with certain properties. The rest of the proof consists of showing that from these properties it follows that $f(c+h)>\alpha$ for all $0 < \lvert h\rvert < \delta.$

Since you worked backward from that conclusion in your before-the-proof work to figure out a good value of $\epsilon$ so that this conclusion would be true, you should be able to reach that conclusion now by working forward from the given statements and other already-proved statements in your proof.

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Implicitly, you assume that $f$ is continuous in $(\alpha,b)$, and hence considering an $\epsilon=l-\alpha>0$ you have the solution that you mentioned. The only crucial assumption to mention here it is the continuity of $f$ and I believe your proof is more than complete!

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  • $\begingroup$ Thanks! It's not mentioned anywhere in the question regarding the continuity. How do I go about proving it? $\endgroup$ – user_9 Aug 17 '19 at 8:32
  • $\begingroup$ If you have some function $f$ s.t. $f:(\alpha,b)\rightarrow$ and you want to show that it is continuous with the definition, then you only need to apply that, for any given $\epsilon>0$, $\exists \delta>0$ s.t. $0<|x-y|<\delta\implies|f(x)-f(y)|<\epsilon$., for $x,y \in(\alpha,b)$...but it seems a little trivial to do such thing. I thought it was implicitly considered. $\endgroup$ – Nav89 Aug 17 '19 at 8:45

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