(4) is continuous at x = c, (provided g (c) = 0). # g Proof We are investigating continuity of (f + g) at x = c. Clearly it is defined at x= c. We have lim(f+g)(x) lim [f(x)+ g(x)] (by definition of f+ g) xc = lim f(x)+lim g(x) (by the theorem on limits) x-c = f(c) + g(c) = (f+ g) (c) xc (as fand g are continuous) (by definition of f+g) Hence, f+g is continuous at x = c. Proofs for the remaining parts are similar and left as an exercise to the reader. 114 MATHEMATICS Rationalised 2023-24 Remarks (i) As a special case of (3) above, iff is a constant function, i.e., f(x) = for some real number , then the function (2. g) defined by (2. g) (x) = g(x) is also continuous. In particular if = 1, the continuity of fimplies continuity of-f (ii) As a special case of (4) above, if f is the constant function f(x) = , then the function g defined by (x)= is also continuous wherever g(x) = 0. In 2 g g(x) particular, the continuity of g implies continuity of g The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not. The following examples illustrate this: Example 16 Prove that every rational function is continuous. blishe Solution Recall that every rational function f is given by p(x) f(x)= q(x)=0 q(x) where p and q are polynomial functions. The domain off is all real numbers except points at which q is zero. Since polynomial functions are continuous (Example 14), fis continuous by (4) of Theorem 1. Example 17 Discuss the continuity of sine function. Solution To see this we use the following facts lim sin x = 0 x0 We have not proved it, but is intuitively clear from the graph of sin x near 0. Now, observe that f(x) = = sin x is defined for every real number. Let c be a real number. Put x = c + h. c we know that h0. Therefore not lim f(x) = lim sin x XC = = xc lim sin(c + h) h0 lim [sin c cos h+ cosc sinh] h0 lim Isin ccosh] + lim [cos c sin hl