# Epsilon-Delta
I write this, because people really don't understand epsilon-delta
. I'm not good at analysis, but at least I understand epsilon-delt
a. So...
## Continuity
Let's start with continuity. Continuous means "no jump". Then what
is a "jump"?
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Suppose, for any d>0, there exists an e>0 s.t. |f(x+-d) - f(x)| >=
e. Then we will say f has a jump at x.
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By slightly changing,
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Suppose, for any d>0, there exists an e>0 s.t. |f(x) - f(x0)| >= e
for every x satisfying |x-x0| < d. Then we will say f has a jump a
t x0.
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Continuous at x means no jump at x. By negating,
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Suppose, for any e>0, there exists an d>0 s.t. for all x, |x-x0| <
d implies |f(x) - f(x0)| < e. Then we will say f is continuous at
x0.
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This is the definition of continuity.
## Limit
If we use p,L instead of x0,f(x0), we get the definition of limit.
This means the limit is what we get by approaching without a jump.
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Suppose, for any e>0, there exists an d>0 s.t. for all x, |x-p| <
d implies |f(x) - L| < e. Then we will say the limit of f, as x ap
proaches p, is L. We write this as
lim_{x->p} f(x) = L
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For x->inf, change (p-d,p+d) with (N0,inf). We can think this as n
o jump at infinity.
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Suppose, for any e>0, there exists an N0 <- N s.t. for all x, x >
N0 implies |f(x) - L| < e. Then we will say the limit of f, as x a
pproaches inf, is L. We write this as
lim_{x->inf} f(x) = L
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It works for a sequence too. Because a_n = f(n).
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Suppose, for any e>0, there exists an N0 <- N s.t. for all n, n >
N0 implies |a_n - L| < e. Then we will say the limit of a_n is L.
We write this as
lim_{n->inf} a_n = L
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