Discuss the continuity of g(x) at x=1

Question

{ (1/x^2-x)+1/x, x<0

g(x)= { [[x-1]], 0<=x<=2

{ √x-2, x>2

Answer 1

A function g(x) is said to be continuous if there are no abrupt changes in value (no discontinuities; no 'holes' or 'jumps) in the interval defined. In other words, for a value at x = c, does f(c) exist, does lim x ==> c f(x) exist and does f(c) = lim x ==> c f(x). For the piecewise function above, we see there are 3 intervals defined: (-inf,0) υ [0,2] υ (2,+inf). We are interested at the continuity at x = 1, so we are only interested in the interval (0,2). Does g(1) exist? Does lim x ==> 1 g(x) exist? f(1) = 1-1=0. So it exists. the lim x ==> 1 f(x) also equals 0. So the function g(x) at x = 1 is continuous.