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Let f(x) = (x * 2^x - x) / (1 - cos x) and g(x) = 2^x * sin(ln2 / 2^x). Which of the following is/are correct? (A) lim_(x->0) f(x) = ln 2 (B) lim_(x->inf) g(x) = ln 4 (C) lim_(x->0) f(x) = ln 4 (D) lim_(x->inf) g(x) = ln 2

1. lim_(x->0) f(x) = ln 2
2. lim_(x->inf) g(x) = ln 4
3. lim_(x->0) f(x) = ln 4
4. lim_(x->inf) g(x) = ln 2

Correct answer: lim_(x->0) f(x) = ln 4

Solution

For lim_(x->0) f(x): f(x) = x*(2^x - 1)/(1 - cosx). As x->0, 2^x - 1 ~ x*ln2 and 1 - cosx ~ x²/2. So f(x) ~ x*(x*ln2)/(x²/2) = 2*ln2 = ln4. So option C is correct, option A is wrong. For lim_(x->inf) g(x): Let t = (ln2)/2^x. As x->inf, 2^x->inf so t->0. g(x) = 2^x * sin(t). Now 2^x = ln2/t, so g(x) = (ln2/t)*sin(t) -> ln2 * (sin t)/t -> ln2 * 1 = ln2. So option D is correct, option B is wrong. Correct options: C and D.

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