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Let f(x) = x² - 4x, g(x) = x + 5, and h(x) = x - 12. Evaluate the limit: lim_(x->7) lim_(n->infinity) ([g(h(x))]^(2n) + 1) / h(f(x)).

1. 4/9
2. 2/9
3. 1/9
4. Limit does not exist

Correct answer: 1/9

Solution

g(h(x)) = g(x-12) = (x-12)+5 = x-7. At x approaching 7, |x-7| < 1 (for x in a neighbourhood of 7), so as n->infinity, (x-7)^(2n) -> 0. The inner limit gives 0+1=1 in the numerator. The denominator: h(f(x)) = f(x)-12 = (x²-4x)-12. At x=7: 49-28-12=9. So the overall limit = 1/9.

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