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**tbyou87** Complete Proof:

Since f is uniformly continuous on (a,b),

for every eps1>0, there exists delt1>0 s.t. x,y ele (a,b) and |x-y|< delt1 => |f(x)-f(y)|< eps1

Since g is uniformly continuous on (a,b),

for every eps2>0, there exists delt2>0 s.t. x,y ele (a,b) and |x-y|< delt2 => |g(x)-f(y)|< eps2

Let eps = eps1 + eps2 >0 and let delt = delt1 + delt2.

Then |x-y|<delt1+delt2<delt => |f(x)-f(y)+g(x)-g(y)|<= |f(x)-f(y)|+|g(x)-g(y)|< eps1+eps2=eps

Thus f+g is uniformly continuous.

Is this ok?