let f and g be functions and let p be a real number in the domain of each. assum
ID: 2942875 • Letter: L
Question
let f and g be functions and let p be a real number in the domain of each. assume that f and g are continuous at p. let cR. prove
a)f+g is continuous at p
b)c•f is continuous at p
c)f•g is continuous at p
Explanation / Answer
Recall that the definition of a function f being continuous at a point a is that the following three properties hold: (1) f(a) exists; (2) lim(x->a) f(x) exists; (3) lim(x->a) f(x) = f(a). Basically, we want that the point exists, the limit exists, and the two are equal. Since we know f and g are continuous at p, we know these three properties are true for f at p and for g at p. a) Show f+g is continuous at p. We show the three properties are still true: (1) We know f(p) and g(p) exist (since f and g are continuous at p). And (f+g)(p) = f(p) + g(p) and therefore also exists. (2) Since lim(x->p) f(x) and lim(x->p) g(x) both exist (since f and g are continuous at p), and lim(x->p)(f+g)(x) = lim(x->p)(f(x)+g(x)) = lim(x->p)f(x) + lim(x->p)g(x) because of the linearity of limits and is therefore the sum of two numbers that we know exist already, lim(x->p)(f+g)(x) exists. (3) In (2), we showed that lim(x->p)(f+g)(x) = lim(x->p)f(x) + lim(x->p)g(x). We know that lim(x->p)f(x) = f(p) and lim(x->p)g(x) = g(p) because f and g are continuous at p, so lim(x->p)f(x) + lim(x->p)g(x) = f(p) + g(p), which by what we showed in (1) equals (f+g)(p). All three properties hold, so f+g is continuous at p. b) Show cf is continuous at p. We show the three properties are still true: (1) Since f is continuous at p, f(p) exists, so cf(p) must also exist. (2) Since lim(x->p)f(x) exists (because f is continuous at p), and lim(x->p)cf(x) = c lim(x->p)f(x) because of the linearity of limits, lim(x->p)cf(x) equals c times something we know exists so must exist as well. (3) We know lim(x->p)f(x) = f(p) because f is continuous at p, so multiplying both sides by c gives cf(p) = c lim(x->p)f(x) = lim(x->p)cf(x) (which we showed in (2)). So cf(p) = lim(x->p)cf(x), and (3) holds for cf at p. All three properties hold, so cf is continuous at p. c) Once again, we show the three properties: (1) Since f(p) and g(p) exist, fg(p) = f(p)g(p) must exist. (2) Since lim(x->p)f(x) and lim(x->p)g(x) exist, and lim(x->p)fg(x) = (lim(x->p)f(x))(lim(x->p)(g(x)) by the properties of limits, it is the product of two numbers that exist and therefore exists. (3) Just as in (a) and (b), we use the simplifications from (1) and (2) above: we know now that fg(p) = f(p)g(p), and we know that lim(x->p)fg(x) = (lim(x->p)f(x))(lim(x->p)(g(x)), which equals f(p)g(p) since (3) is true for f and g at p. And that proves (3) for fg(p). All three properties hold, so fg is continuous at p. d) Same thing: (1) f(p)/g(p) is the quotient of two numbers that exist (because of the continuity of f and g at p) and therefore it must exist as long as g(p) does not equal 0, which is specified. (2) We have a quotient rule for limits as well, which gives lim(x->p)f(x)/g(x) = [lim(x->p)f(x)]/[lim(x->p)g(x)]. We know these both exist because f and g are continuous at p. We also know lim(x->p)g(x) does not equal 0, because according to (3), since g is continuous at p we must have g(p) = lim(x->p)g(x) and g(p) does not equal 0 according to the problem. So lim(x->p)f(x)/g(x) exists. (3) Since lim(x->p)f(x)/g(x) = [lim(x->p)f(x)]/[lim(x->p)g(x)] and the continuity of f and g at p tells us this equals f(p)/g(p), (3) also holds for f(x)/g(x) at p. All three properties hold, so f(x)/g(x) is continuous at p. e) Again we prove the three properties: (1) Since f is continuous at p, f(p) exists. Since g is continuous at f(p), g(f(p)) exists. This is the same as saying gof exists at p. (2) We have a theorem about limits that says that if lim(x->a)f(x) = b and lim(x->b)g(x) = L, then lim(x->a)g(f(x)) = L. Since f is continuous at p, we have lim(x->p)f(x) = f(p) by (3). Since g is continuous at f(p), we know lim(x->f(p))g(x) exists; let it equal L. Then by our theorem, we have lim(x->p)g(f(x)) = L, so it exists as well. (3) Again by our theorem, lim(x->p)g(f(x)) = lim(x->f(p))g(x). We know lim(x->f(p))g(x) = g(f(p)) because of continuity of g at f(p), so lim(x->p)g(f(x)) = g(f(p)), and (3) holds. All three properties hold, so gof is continuous at p. Notice that what we have proven here are basic theorems about continuity -- they will allow you to combine functions in all of these ways and ensure that the resulting function is still continuous.
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