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What I question is the limit function $\lim$ which we use to find to limit of a function when its parameter comes very close to some value. Also, according to the definition of limit that is for any arbitrary number $\epsilon >0$, there exist a number $\delta$ such that for $0 <|x-a|<\delta$ then $0<|f(x)-f(a)|<\epsilon$.

So to prove the limit of $f(x)$ as $x \to a$ is $L$, I must prove that there exist a number $\delta$ such that for $0 <|x-a|<\delta$ then $0<|f(x)-L|<\epsilon$.

What I don't know is why we must still use that way to prove the method. Isn't the $\lim$ function always right with its rule because we use it to find the limit?

For example, the limit of $f(x) = x-3$ as $x$ approach $3$ will be zero. We can use the $\lim$ function to solve it.

So what is the point to prove it by using the $\epsilon, \delta$ if using the $\lim$ function will get us the right result?

I mean, if using $\lim$ function can get wrong result then people should not use it. We use it because if we follow all its rules the result we get is always right, so why must we prove the limit of $f(x) = x-3$ as $x$ approach $3$ will be zero again by proving that there exist a number $\delta$ such that for $0 <|x-3|<\delta$ then $0<|f(x)-0|<\epsilon$?

Why isn't going straight to the result using $\lim$ enough, but still having to use $\delta, \epsilon$ to prove it again?

Peter Mortensen
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aukxn
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    It's hard to follow all the above (using spacing and line-spacing more often is greatly adviced), but you seem to believe that evaluating limits of functions is always substitution. This is far from being true, although it is true if the function's continuous at the point where $;x;$ approaches. – Timbuc Oct 15 '14 at 13:30
  • I think I've deciphered your question: you're saying that you've learned a list of rules to obtain the limit of a function at a point, and you're wondering why you have to revert to using the definition of a limit to prove existence if said rules are correct. One possible answer is that there are nasty expressions for which the rules you've learned might not suffice in proving what the limit is, however you might be otherwise capable of proving that it does exist. But that's a bit of a cop out—I'm going to say the real reason is because your teacher wants you to learn the definition. – G. H. Faust Oct 15 '14 at 15:33

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"Going straight to the result" means exactly that somebody used $\epsilon$ and $\delta$ to provide you with the "straight" result. Why can you use $\lim_{x \to 0} (x+3) =3$? Because you, or someone else, proved that for every $\epsilon>0$ there exists $\delta >0$ such that...

And how can you go straight to the result to compute $$ \lim_{x \to 0} \frac{\sin x}{x}? $$

Siminore
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    you use young formula : sin(x) = x + o(x) :D You're right though – mvggz Oct 15 '14 at 13:41
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    I always told my students that the reason for the $\delta-\epsilon$ proofs was to derive those (more general) theorems so that we could avoid using the $\delta-\epsilon$ proofs every time we needed to evalulate a (specific) limit. I then spent some classroom time going over the proofs of those more general theorems to show how that worked. One benefit of that method was that they now new how to attack the more specific limit problems. – Steven Alexis Gregory Mar 18 '18 at 11:14