Chapter 1: Limits
Section 1.6: Continuity
Example 1.6.6
If fx=x−1x<15x≥1 and gx=2 xx<12−5 xx≥1, show that the rule for f+gx=3 x−1x<17−5 xx≥1 is continuous at x=1 even though neither f nor g is continuous at x=1.
Solution
In Figure 1.6.6(a), fx is graphed in blue, gx is graphed in green, and their sum, f+gx, is graphed in red. From this Figure, it should be clear that both f and g have jump discontinuities at x=1, but that the rule for the sum does not.
To be very clear: The rule for the sum does not have a discontinuity at x=1, but the sum of f and g is defined as a function only on the domain common to both f and g. This common domain does not include x=1, so the function that is their sum cannot have x=1 in its domain. But looking at just the rule for the sum, independent of whence it came, that rule can be assigned the largest possible domain for which it is defined, and that largest domain is the set of all the real numbers.
unassign('f','g'): F:=piecewise(x<1,x-1,5): G:=piecewise(x<1,2*x,2-5*x): plot([F,G,F+G], x=-2..3, discont=true, color=[blue,green,red],legend = [typeset(f(x)),typeset(g(x)),typeset(f(x)+g(x))]);
Figure 1.6.6(a) Graphs of f, g, and f+g
Show f is discontinuous at x=1
Show g is discontinuous at x=1
fx=x−1x<15x≥1→assign as functionf
gx=2 xx<12−5 xx≥1→assign as functiong
f1 = 5
limx→1−fx = 0 but limx→1+fx = 5
g1 = −3
limx→1−gx = 2 but limx→1+gx = −3
Show f+gx is continuous at x=1
Evaluate f+g at x=2 Context Panel: Evaluate and Display Inline
f+g1 = 2
Expression palette: Limit operator Context Panel: Evaluate and Display Inline
limx→1f+gx = 2
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