A doubt in the rigorous definition of limits.

I studied the definition of limits today, and I think I mostly understood it, but I have a little doubt. In the definition:

$f(x)$ is defined on some open interval containing $a$, except at possibly $a$. So, $\lim_{x\to a} f(x) = L$ if and only if for every number $\varepsilon>0$, there exists a corresponding number $\delta>0$ such that: $$\text{If } 0<|x-a|<\delta \text{ then } |f(x)-L|<\varepsilon$$

First of all, if you see any mistake in my definition or understanding, please tell. So my question is, why can't we write $0<|x-a|\leq\delta$ and $|f(x)-L|\leq\varepsilon$ in the last line? What kinds of problems can arise from this? If the definitions are equivalent, can you explain/prove me why?

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Is epsilon not delta in the image. –  rlartiga Apr 17 at 14:11
Should be $\forall \epsilon >0 \exists \delta >0: 0<|x-c|<\delta \implies |f(x)-f(c)|<\epsilon$ –  Sujaan Kunalan Apr 17 at 14:12
You can express the limits using $\le$ instead of $<$ as long as the $\epsilon,\delta >0$. –  copper.hat Apr 17 at 14:16
Yes, all in order. I don't know really how explain it, but the two definition are equivalent (If f is continous: the preimage of a closed set is closed if and only if the preimage of an open set is open) –  rlartiga Apr 17 at 14:16
You should also specify what function the limit is acting on. $\lim_{x\to a}=L$ has no meaning. I think you meant $\lim_{x\to a} f(x)=L$ –  Sujaan Kunalan Apr 17 at 14:17

First assume: $\varepsilon>0$, there exists a corresponding number $\delta>0$ such that: $$\text{If } 0<|x-a|<\delta \text{ then } |f(x)-L|<\varepsilon$$

For that definition take $\varepsilon=\frac{\epsilon}{2}$, then exist $\delta$ such: $$\text{If } 0<|x-a|<\delta \text{ then } |f(x)-L|<\frac{\epsilon}{2}$$ Then you can see: $$0<|x-a|\leq\frac{\delta}{2}\Rightarrow 0<|x-a|<\delta\Rightarrow |f(x)-L|<\frac{\epsilon}{2}\leq{\epsilon}$$ Then take $\delta'=\frac{\delta}{2}$ and you have your definition with less or equal.

The other side is analogous

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1. if you want $\leq \varepsilon$, just choose $2\varepsilon$ in the definition with $<$;
2. if you want $\leq \delta$, just choose $\frac{\delta}{2}$ in the definition with $<$.

In the first case, you exploit the words "for all", in the second the words "for some".

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The definition says, for every $\epsilon>0$ there exists a $\delta>0$ which is the crucial point to understand. Here is how I explain:

When $x$ tends to $a$, we are approaching a point $a$ where the value of the function is $L$, while approaching, at any arbitrary $x$ the value of the function is $f(x$). We can say we are approaching the limit $L$ only if the difference, say $df$, between $f(x)$ and $L$ decreases as the difference, say $dx$, between $x$ and $a$ decreases. This decrease in $dx$ ensures that $x$ is approaching towards $a$ and the decrease in $df$ ensures that $f(x)$ is approaching towards $L$. Which means while approaching, the differences must continuously decrease and there always is an "$\epsilon$" above $df$ and a "$\delta$" above $dx$ which ensures we are approaching the limit. Since $\delta$ and $\epsilon$ can take any positive infinitesimal value, the conditions $0<|x-a|<\delta$ and $|f(x)-L|<\epsilon$ tells that $|a-x|$ and $|f(x)-L|$ can take even much smaller positive values than $\delta$ and $\epsilon$ respectively, ensuring the approach to be very close to $L$

If we write $0<|x-a|\leq\delta$ and $|f(x)-L|\leq\epsilon$, the equality conditions gives $\epsilon=|f(x)-L|$ and $\delta=|x-a|$, which can be satisfied anywhere on the curve and the equality in either of them or in both will not ensure that the differences are still smaller than whatever value $\epsilon$ and $\delta$ takes and hence will not ensure that they are decreasing or we are approaching the limit. When we say "less than or equal to" any of the condition is satisfied, but when we say "less than" the only condition satisfied is $|x-a|$ and $|f(x)-L|$ will always remain smaller than $\delta$ and $\epsilon$ respectively and when the definition says for every $\epsilon>0$ there exists $\delta>0$ such that

$0<|x-a|<\delta\space\space\space\space$ for $\space\space|f(x)-L|<\epsilon\space$ it means that whatever value $\epsilon$ and $\delta$ takes while approaching, $|x-a|$ and $|f(x)-L|$ are still smaller than them.

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+1 Nice :) ${}{}$ –  Sawarnik Apr 17 at 17:52