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prove the statement using the epsilon, delta definition of a limit. limit x to a of c=c
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let delta=d let epsilon =e so let's think f(x)=c L=c so |f(x)-L|<e |c-c|<e |0|<e 0<e --------------------- it doesn't matter what we choose delta to be here choose d= ? .so whenever 0<|x-a|<d, then |f(x)-L|=|c-c|=0d=0?=0<e
what about lim of x to 1 to (2+4x)/3=2 Would that be the same format? f(x)= (2+4x)/3 and L=2
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