Prove that if the functions f + g: R -> R and g: R -> R are continuous, then so

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I assume f+g is defined by (f+g)(x) = f(x) + g(x). If f+g and g are continuous on R, then they are each continuous at every real number. Let r be an arbitrary real number. Then by the continuity of f+g, (f+g)(r) = lim(x->r)((f+g)(x)). But (f+g)(r) = f(r) + g(r) and lim(x->r)((f+g)(x)) = (lim(x->r)(f(x)) + lim(x->r)(g(x)), so f(r) + g(r) = (lim(x->r)(f(x))) + (lim(x->r)(g(x))). Because g is continuous, g(r) = lim(x->r)(g(x)). So we can cancel out g(r) on the left and lim(x->r)(g(x)) on the right of the first equation, yielding f(r) = lim(x->r)(f(x)). But this is just the definition of continuity for f at r. Since we assumed that r is any arbitrary real number, f is continuous at all points in R. Therefore f: R -> R is continuous.

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