Solução_Calculo_Stewart_6e

F.
56 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
4
11. lim f(x) = lim x +2x
3 4 = lim x +2 lim
x→−1 x→−1
x→−1 x→−1 x3 = −1+2(−1) 34 =(−3) 4 =81=f(−1).
By the definition of continuity, f is continuous at a = −1.
13. For a>2,wehave
2x +3
lim (2x +3)
lim f(x)= lim
x→a x→a x − 2 = x→a
lim (x − 2) [Limit Law 5]
x→a
2 lim x + lim 3
x→a x→a
=
lim x − lim 2
x→a x→a
=
2a +3
a − 2
= f(a)
TX.10
[1,2,and3]
[7 and 8]
Thus, f is continuous at x = a for every a in (2, ∞);thatis,f is continuous on (2, ∞).
15. f(x) =ln|x − 2| is discontinuous at 2 since f(2) = ln 0 is not defined.
17. f(x) =
e
x
if x