Answer :
L1 and L2 are limits of f at xo, thus |L1-L2|<ε implies L1 = L2 by the definition of limit.
If L1 and L2 are limits of f at xo, then for every ε > 0, there exist δ1, δ2 > 0 such that 0 < | x - xo | < δ1, and 0 < | x - xo | < δ2 implies | f(x) - L1 | < ε/2 and | f(x) - L2 | < ε/2, respectively.
Therefore, for any ε > 0, there is a δ = min
{δ1, δ2} > 0, such that 0 < | x - xo | < δ implies | f(x) - L1 | < ε/2 and | f(x) - L2 | < ε/2.
Thus, | L1 - L2 | ≤ | L1 - f(x) | + | f(x) - L2 | < ε/2 + ε/2 = ε.
Since ε can be made arbitrarily small, it follows that L1 = L2.
L1 and L2 are limits of f at xo, thus |L1-L2|<ε implies L1 = L2 by the definition of limit.
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