Verified answer

Given the condition, the function f - g is positive, or zero, everywhere in D. That is f - g lies in the set [0, infinity), which is closed. Therefore its limit must lie within [0, infinity). We know that f(x) - g(x) ---> K - L by the algebra of limits, so K - L lies in [0, infinity), i.e.

K - L >= 0

K >= L

Consider the function h(x) = f(x) - g(x)

Then, h(x) ≥ 0 or all x ∈ D and lim(x-->a) h(x) = K-L.

Now assume K-L <0

Choose e = -(K-L)/2.

Then, by the definition of the limit one can find delta > 0 such that for any x ∈ D in (a - delta, a + delta) (since a is accumulation point, such x always exists)

K-L - e < h(x) < K-L + e, or equivalently, (3/2)(K-L) < h(x) < (1/2)(K-L)

And since K-L < 0 ==> h(x) < 0 which creates a contradiction (since h ≥ 0).

Hence, K-L≥0.

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