Answer to Question #750 in Calculus for Luke
equal to x.
(a) limx →n ⌊x⌋ and
(b) limx→n+1/2 ⌊x⌋ exist,
when n ∈ Z. If the limit exists give it.
lim(x->n-)= n-1 from the left side of n;
lim(x->n+)= n from the right side of n.
n+1/2 is always less than n+1, thus
lim(x-> n+1/2) = n.
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