As x tends to +-infinity, (x-3) also tends to +-infinity, 1/(x-3) tends to 0.
So the value in question is 2. (a)
When x is large and positive, (x-3)>0, 1/(x-3) > 0, so f(x) > 2 (b)
When x is large in magnitude and negative, 1/(x-3) < 0, so f(x) < 2 (c)
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