What if we change the def of limit as following In definition of limits why can't we have " there exist delta for all epsilon" instead of  " for all epsilon there exist delta"
 A: "For any $\epsilon$, there exists a $\delta$" means that whenever you get an $\epsilon$, it's possible to find some $\delta$ that makes it work.
"There exists a $\delta$ such that for any $\epsilon$" means that there is one, single $\delta$ that works no matter what $\epsilon$ might be.
An analogy might do: Take the statement "For any man, there is a woman that is meant for him". It is not an outrageous statement. Some might contest it, but the idea has been there for centuries, and many people will defend its truth to their dying breath.
The statement "There is a woman, such that for any man, she is meant for him", on the other hand, implies that there is one woman somewhere that is everyone's future wife (poor girl). I think you will have to look hard to find anyone who actually believes this.
A: But a necessary (though not sufficient) condition for that to happen is that $f$ must be uniformly continuous, and not just continuous; otherwise,$\delta$ is a function of $\epsilon $
A: Because the idea of the limit is that "we can make $f(x)$ arbitrarily close to $L$, by making $x$ sufficiently close to $a$". In other words, instead of analyzing the domain and seeing what happens in the range, we analyze the range, and try to modify the domain appropriately. So we say "for all $\epsilon$ (Range) there exists a $\delta$ (Domain)" instead of vice versa.
A: I'll work a similar concept: Continuity.
If we have in the definition "there exists a delta such that for all epsilon", then a lot of intuitively "continuous" functions (those whose graph can be drawn without raising the pencil) wouldn't be continuous. 
For instance, take the identity function $f(x)=x$ defined on $\mathbb R$, then $f$ is continuous at $0$ in the intuitive way but not with the "new" definition since it's not true that there exists a $\delta>0$ such that for every $\varepsilon>0$ (whatever you choose), if $|x|<\delta$ then $|f(x)|=|x|<\varepsilon$ (if such a $\delta$ exists, then take $x=\varepsilon=\delta/2$, we have $|x|<\delta$ but $|f(x)|=|x|\nless \varepsilon$). 
A: ‘There exists a $\delta>0$ such that for all $\varepsilon>0$, if $\,\lvert x-x_0\rvert <\delta$, then $\,\lvert f(x)-\ell\rvert<\varepsilon$’
implies that $f(x)-\ell=0$, i. e. $f(x)=\ell\,$ as soon as $x\in (x_0-\delta,x_0+\delta)$.
