Is the Validity of the Proof in Question?

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SUMMARY

The discussion centers on proving the limit definition: the limit of f(x) as x approaches c equals L if and only if the limit of |f(x) - L| as x approaches c equals 0. A participant attempts to simplify the proof by asserting that if the limit of |f(x) - L| equals 0, then f(x) must equal L. However, this reasoning is challenged, emphasizing that f(x) is a function of x, while L is a constant. The correct approach involves applying the formal definition of limits, which requires the use of epsilon-delta arguments.

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semidevil
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I need to prove that the

limit of as x goes to c, f(x) = L iff limit as x goes to c |f(x) - L| = 0.

I"m still new to this limit concept, but here is the only thing I can think of.

So I break it down into 2 parts since it is an IFF problem. Prove ---> and then prove <----

I mean, can't I Just say that if the limit as x goes to c of |f(x) - L|= 0, then f(x) = L(because, if f(x) is not equal to L, then it wouldn't equal 0...like, a - a = 0, but b - a is not equal 0).. So I can put L to the other side, and proof done. And same the other way.

Is this a valid proof?

or is there more to it?
 
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semidevil said:
I mean, can't I Just say that if the limit as x goes to c of |f(x) - L|= 0, then f(x) = L(because, if f(x) is not equal to L, then it wouldn't equal 0...like, a - a = 0, but b - a is not equal 0)..

This is nonsense. NO, f(x) does not have to be equal to L: f(x) is a function of x and L is a constant. lim(x->0) x2= 0 but it certainly is not true that x2= 0 !

Remember the definition of "lim (x->ac) f(x)= L": Given any &epsilon;> 0, there exist &delta;> 0 such that if |x-c|< &delta;, then |f(x)-L|< &epsilon;

Now apply that same definition to "lim(x->c) f(x)-L= 0". Replace f(x) by f(x)-L and replace L by 0. What happens?
 

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