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

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

• A. f(g(a))
• B. g(a)
• C. f(a)
• D. a

Answer: (C)

Key idea: when the inner function g(x) approaches a as x approaches a, and the outer function f is continuous at a, you can pass the limit through the outer function. This means the limit of the composition is f evaluated at the inner limit: lim_{x→a} f(g(x)) = f(lim_{x→a} g(x)) = f(a). Intuition: near x = a, g(x) is getting arbitrarily close to a. Because f is continuous at a, values of f near a stay close to f(a). So as x gets close to a, f(g(x)) gets arbitrarily close to f(a), driving the limit to f(a). Why the others don’t fit: f(g(a)) would use the exact inner value at x = a, not the limiting behavior as x → a. The limit depends on how g(x) behaves near a, not necessarily on the value at a. The limit is a value of f, not the number a itself or g(a).

Key idea: when the inner function g(x) approaches a as x approaches a, and the outer function f is continuous at a, you can pass the limit through the outer function. This means the limit of the composition is f evaluated at the inner limit: lim_{x→a} f(g(x)) = f(lim_{x→a} g(x)) = f(a).

Intuition: near x = a, g(x) is getting arbitrarily close to a. Because f is continuous at a, values of f near a stay close to f(a). So as x gets close to a, f(g(x)) gets arbitrarily close to f(a), driving the limit to f(a).

Why the others don’t fit: f(g(a)) would use the exact inner value at x = a, not the limiting behavior as x → a. The limit depends on how g(x) behaves near a, not necessarily on the value at a. The limit is a value of f, not the number a itself or g(a).