Assume that f,g : R {1} R are two functions such that lim x1 f(x) = 2 and lim x1 g(x) = 3. We'll show that lim x1 (4 f(x) +g(x) 2 ) = 17. Let be positive. Assume first that 1. We'll deal with the case > 1 later. There exists a positive 1 such that, for all x with 0 < |x1| < 1 we have | f(x)2| < 11 . There exists a positive 2 such that, for all x with 0 < |x1| < 2, we have |g(x)3| < 11 . Observe that for all x with 0 < |x1| < 2, |g(x) +3| = |g(x)3+6| |g(x)3|+6 11 +6 1 11 +6 7. Let = min{1,2}. For all x with 0 < |x1| < we have (4 f(x) +g(x) 2 )17 = |4(f(x)2) + (g(x) +3)(g(x)3)| = 4| f(x)2|+|g(x) +3| |g(x)3| < 4 11 +7 11 = . This completes the argument in the case 1. Assume now that > 1. By our work above, there exists a positive such that, for all x with 0 < |x1| < , (4 f(x) +g(x) 2 )17 < 1, which implies (4 f(x) +g(x) 2 )17 < . >1<1