If lim_{x->a} g(x) = a and f is continuous at a, what is lim_{x->a} [f(g(x))]^2?

If lim_{x->a} g(x) = a and f is continuous at a, what is lim_{x->a} [f(g(x))]^2?

• A. [f(a)]^2
• B. [f(g(a))]^2
• C. [a]^2
• D. No limit

Answer: (A)

The key idea is handling a composition of limits with a continuous outer function. Since g(x) approaches a as x approaches a, the inner input to f gets arbitrarily close to a. because f is continuous at a, f(g(x)) then approaches f(a). So lim x→a f(g(x)) = f(a). The squaring operation is also continuous, so taking the limit after squaring gives [lim x→a f(g(x))]^2 = [f(a)]^2. Therefore, the limit is [f(a)]^2. This relies on f being defined and continuous at a; otherwise the limit could behave differently.

The key idea is handling a composition of limits with a continuous outer function. Since g(x) approaches a as x approaches a, the inner input to f gets arbitrarily close to a. because f is continuous at a, f(g(x)) then approaches f(a). So lim x→a f(g(x)) = f(a). The squaring operation is also continuous, so taking the limit after squaring gives [lim x→a f(g(x))]^2 = [f(a)]^2. Therefore, the limit is [f(a)]^2. This relies on f being defined and continuous at a; otherwise the limit could behave differently.