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Chapter 2
Differentiation f ) Since lim g(x) = g(−2), then the function is not x→−2
Exercise Set 2.1
continuous at x = −2
1. The function is not continuous at x = 1 since the limit from the left of x = 1 is not equal to the limit from the right of x = 1 and therefore the limit of the function at x = 1 does not exist.
7.
x→1
from the left we notice that the y-value is approaching a value of 2. Thus, lim h(x) = 2. Since lim h(x) = x→1−
2. The function is not continuous at x = −2 since the limit from the left of x = −2 is not equal to the limit from the right of x = −2 and therefore the limit of the function at x = −2 does not exist.
b) Reading the value from the graph h(1) = 2. c) Since the lim h(x) = 2 and = h(1) = 2 then h(x) is x→1 continuous at x = 1. d) As we approach the x-value of -2 from the right we notice that the y-value is approaching a value of 3. Thus, lim h(x) = 0. As we approach the x-value + x→−2
of -2 from the left we notice that the y-value is approaching a value of 0. Thus, lim − h(x) = 0. Since
a) As we approach the x-value of 1 from the right we notice that the y-value is approaching a value of 1. Thus, lim+ f (x) = −1. As we approach the x-
lim h(x) = +
x→−2
x→1
x→1−
x→−2
tinuous at x = −2. 8.
x→1
d)
x→−2
x→−2
f ) Since lim t(x) does not exist, then the function is
x→−2
x→−2
e) Reading the value from the graph f (−2) = 3.
6.
a) lim+ g(x) = −2, x→1
lim f (x) = −2
lim− g(x) = −2,
x→1
not continuous at x = −2 9.
therefore
x→1
lim g(x) = 4,
does not exist
x→1
b) As we approach the x value of 1 from the left, we find that the y value is approching 3. Thus lim f (x) = 3 c) Since
c) Since lim g(x) = −1 = g(1) = −2, then g(x) is conx→1 tinuous at x = 1 x→−2+
a) As we approach the x value of 1 from the right we find that the y value is approching 3. Thus lim+ f (x) = 3
x→1−
b) g(1) = −2
d)
lim t(x) = undefined,
x→−2−
e) t(−2) = undefined
lim f (x) = 3 then lim f (x) = 3.
x→−2
lim t(x) = undefined,
x→−2+
thus lim t(x) does not exists
x→−2
tinuous at x = −2.
therefore
c) Since lim t(x) = t(1) ≈ 0.25, then t(x) is continuous x→1 at x = 1
of -2 from the left we notice that the y-value is approaching a value of 3. Thus, lim − f (x) = 3. Since
f ) Since lim f (x) = 3 and f (−2) = 3, then f (x) is con-
lim t(x) ≈ 0.25,
x→1−
b) t(1) =≈ 0.25
d) As we approach the x-value of -2 from the right we notice that the y-value is approaching a value of 3. Thus, lim + f (x) = 3. As we approach the x-value
x→−2−
x→1
lim t(x) ≈ 0.25
continuous at x = 1.
lim f (x) =
a) lim+ t(x) ≈ 0.25, x→1
c) Since the lim− f (x) does not exist, then f (x) is not
x→−2+
x→−2
x→−2
f ) Since lim h(x) = 0 and h(−2) = 0, then h(x) is con-
x→1−
b) Reading the value from the graph f (1) = −1.
x→−2
lim − h(x) = 0 then lim h(x) = 0.
e) Reading the value from the graph h(−2) = 0.
x→1
value of 1 from the left we notice that the y-value is approaching a value of 2. Thus, lim f (x) = 2. Since lim f (x) = lim f (x) then lim f (x) does not exist.
x→1
x→1
4. The function is not continout at x = −2 since the value of the function at x = −2 is undefined.
x→1+
x→1+
lim− h(x) = 2 then lim h(x) = 2.
3. The function is continuous at every point in the given plot. Note that the graph can be traced without a jump from one point to another.
5.
a) As we approach the x-value of 1 from the right we notice that the y-value is approaching a value of 2. Thus, lim+ h(x) = 2. As we approach the x-value of 1
lim g(x) = −3, thus
x→−2−
lim g(x)
x→−2
e) g(−2) = −3
x→1
lim f (x) = 3
lim f (x) = 3 then
x→1−
x→1
d) From the given conditions f (1) = 2 e) f (x) is not continuous at x = 1 since lim f (x) = f (1) x→3
f ) f (x) is continuous at x = 2 since lim f (x) = 2 = f (2)
x→2−
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lim+ f (x) = 3 and
lim f (x) =
x→2+