If lim_{x->a} f(x) = L and lim_{x->a} g(x) = M, what is lim_{x->a} (f(x) + g(x))?

If lim_{x->a} f(x) = L and lim_{x->a} g(x) = M, what is lim_{x->a} (f(x) + g(x))?

• A. L-M
• B. LM
• C. L+M
• D. 0

Answer: (C)

The key idea is that limits distribute over addition. If f(x) approaches L and g(x) approaches M as x approaches a, and both limits are finite, then the sum f(x) + g(x) approaches L + M. This follows from the idea that you can make f(x) close to L and g(x) close to M simultaneously: for any small ε, you can ensure |f(x) − L| < ε/2 and |g(x) − M| < ε/2 when x is sufficiently near a, which gives |(f(x) + g(x)) − (L + M)| ≤ |f(x) − L| + |g(x) − M| < ε. So the limit of the sum is L + M. The other expressions would only hold in special cases (for example, a difference instead of a sum, or a product, or zero under particular conditions) but the general rule for finite limits is to add them.

The key idea is that limits distribute over addition. If f(x) approaches L and g(x) approaches M as x approaches a, and both limits are finite, then the sum f(x) + g(x) approaches L + M.

This follows from the idea that you can make f(x) close to L and g(x) close to M simultaneously: for any small ε, you can ensure |f(x) − L| < ε/2 and |g(x) − M| < ε/2 when x is sufficiently near a, which gives |(f(x) + g(x)) − (L + M)| ≤ |f(x) − L| + |g(x) − M| < ε.

So the limit of the sum is L + M. The other expressions would only hold in special cases (for example, a difference instead of a sum, or a product, or zero under particular conditions) but the general rule for finite limits is to add them.