CHAPTER 1 LIMITS
SECTION 1.1 Introduction to Limits
Problem Set 1.1, Number 29 – 58.
- For the function f graphed in Figure 11, find the indicated limit or function value, or state that it does not exist.
- Follow the directions of Problem 29 for the function f graphed in Figure 12.
- For the function f graphed in Figure 13, find the indicated limit or function value, or state that it does not exist.
- For the function f graphed in Figure 14, find the indicated limit or function value, or state that it does not exist.
- Sketch the graph of
- Sketch the graph of
- Sketch the graph of f(x) = x –〚x〛; then find each of the following or state that does not exist.
- Follow the directions of Problem 35 for f(x) = x/|x|.
- Find limx → 1 (x2 – 1)/|x – 1| or state that it does not exist.
- Evaluate limx → 0 [√(x + 2) – √2]/x. Hint: Rationalize the numerator by multiplying the numerator and denominator by √(x + 2) + √2.
- Let
- Sketch, as best you can, the graph of a function f that satisfies all the following conditions.
- Let
- The function f(x) = x2 had been carefully graphed, but during the night a mysterious visitor changed the values of f at a million different places. Does this affect the value of limx → a f(x) at any a? Explain.
- Find each of the following limits or state that it does not exist.
- Find each of the following limits or state that it does not exist.
- Find each of the following limits or state that it does not exist.
- Find each of the following limits or state that it does not exist.
- lim√xx → 0
- limxxx → 0+
- lim√|x|x → 0
- lim|x|xx → 0
- lim(sin 2x) / 4xx → 0
- lim(sin 5x) / 3xx → 0
- limcos (1 / x)x → 0
- limx cos (1 / x)x → 0
- limx → 1x3 – 1√(2x + 2) – 2
- limx → 0x sin 2xsin (x2)
- limx → 2–x2 – x – 2|x – 2|
- limx → 1+21 + 21/(x – 1)
- Since calculus software packages find limx → a f(x) by sampling a few values of f(x) for x near a, they can be fooled. Find a function f for which limx → a f(x) fails to exist but for which your sortware gives a valus for the limit.
|
(a)
|
lim
|
f(x)
|
|
|
x → –3
|
|
||
|
(b)
|
f(–3)
|
||
|
(c)
|
f(–1)
|
||
|
(d)
|
lim
|
f(x)
|
|
|
x → –1
|
|
||
|
(e)
|
f(1)
|
||
|
(f)
|
lim
|
f(x)
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|
|
x → 1
|
|
||
|
(g)
|
lim
|
f(x)
|
|
|
x → 1–
|
|
||
|
(h)
|
lim
|
f(x)
|
|
|
x → 1+
|
|
||
|
(i)
|
lim
|
f(x)
|
|
|
x → –1+
|
|
||
|
❤
|
PEMBAHASAN:
|
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
x → –3
|
f(x) does not
exist.
|
|||||
|
(b)
|
f(–3) = 1
|
||||||
|
(c)
|
f(–1) = 1
|
||||||
|
(d)
|
lim
x → –1
|
f(x) = 2
|
|||||
|
(e)
|
f(1) = 1
|
||||||
|
(f)
|
lim
x → 1
|
f(x) does not
exist.
|
|||||
|
(g)
|
lim
x → 1–
|
f(x) = 1
|
|||||
|
(h)
|
lim
x → 1+
|
f(x) does not
exist.
|
|||||
|
(i)
|
lim
x → –1+
|
f(x) = 2
|
|||||
|
(a)
|
f(–3)
|
||
|
(b)
|
f(3)
|
||
|
(c)
|
lim
|
f(x)
|
|
|
x → –3–
|
|
||
|
(d)
|
lim
|
f(x)
|
|
|
x → –3+
|
|
||
|
(e)
|
lim
|
f(x)
|
|
|
x → –3
|
|
||
|
(f)
|
lim
|
f(x)
|
|
|
x → 3+
|
|
||
|
❤
|
PEMBAHASAN:
|
|
(a)
|
f(–3) = 2
|
|||
|
(b)
|
f(3) is
undefined.
|
|||
|
(c)
|
lim
x → –3–
|
f(x) = 2
|
||
|
|
||||
|
(d)
|
lim
x → –3+
|
f(x) = 4
|
||
|
|
||||
|
(e)
|
lim
x → –3
|
f(x) does
not exist.
|
||
|
|
||||
|
(f)
|
lim
x → 3+
|
f(x) does
not exist.
|
||
|
|
||||
|
(a)
|
lim
|
f(x)
|
|
|
x → –1–
|
|
||
|
(b)
|
lim
|
f(x)
|
|
|
x → –1+
|
|
||
|
(c)
|
lim
|
f(x)
|
|
|
x → –1
|
|
||
|
(d)
|
f(–1)
|
||
|
(e)
|
lim
|
f(x)
|
|
|
x → 1
|
|
||
|
(f)
|
f(1)
|
||
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
x → –1–
|
f(x) = –2
|
|
|
|
|||
|
(b)
|
lim
x → –1+
|
f(x) = –2
|
|
|
|
|||
|
(c)
|
lim
x → –1
|
f(x) = –2
|
|
|
|
|||
|
(d)
|
f(–1) = –2
|
||
|
(e)
|
lim
x → 1
|
f(x) = 0
|
|
|
|
|||
|
(f)
|
f(1) = 0
|
||
|
f(x)
|
=
|
{
|
–x
|
if
|
x < 0
|
|
x
|
if
|
0 ≤ x < 1
|
|||
|
1 + x
|
if
|
x ≥ 1
|
Then find each of the following or state that it does not exist.
|
(a)
|
lim
|
f(x)
|
|
|
x → 0
|
|
||
|
(b)
|
lim
|
f(x)
|
|
|
x → 1
|
|
||
|
(c)
|
f(1)
|
||
|
(d)
|
lim
|
f(x)
|
|
|
x → 1+
|
|
||
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
x → 0
|
f(x) = 0
|
||
|
(b)
|
lim
x → 1
|
f(x) does
not exist.
|
||
|
(c)
|
f(1) = 2
|
|||
|
(d)
|
lim
x → 1+
|
f(x) = 2
|
||
|
g(x)
|
=
|
{
|
–x
+ 1
|
if
|
x < 1
|
|
x – 1
|
if
|
1 < x < 2
|
|||
|
5 – x2
|
if
|
x ≥ 2
|
Then find each of the following or state that it does not exist.
|
(a)
|
lim
|
g(x)
|
|
|
x → 1
|
|
||
|
(b)
|
g(1)
|
||
|
(c)
|
lim
|
g(x)
|
|
|
x → 2
|
|
||
|
(d)
|
lim
|
g(x)
|
|
|
x → 2+
|
|
||
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
x → 1
|
g(x) = 0
|
|
|
(b)
|
g(1) does not
exist.
|
||
|
(c)
|
lim
x → 2
|
g(x) = 1
|
|
|
(d)
|
lim
x → 2+
|
g(x) = 1
|
|
|
BACA JUGA:
|
|
|
(a)
|
f(0)
|
||
|
(b)
|
lim
|
f(x)
|
|
|
x → 0
|
|
||
|
(c)
|
lim
|
f(x)
|
|
|
x → 0–
|
|
||
|
(d)
|
lim
|
f(x)
|
|
|
x → 1/2
|
|
||
|
❤
|
PEMBAHASAN:
|
f(x) = x –〚x〛
|
(a)
|
f(0) = 0
|
|||
|
(b)
|
lim
x → 0
|
f(x) does
not exist.
|
||
|
(c)
|
lim
x → 0–
|
f(x) = 1
|
||
|
(d)
|
lim
x → 1/2
|
f(x) = 1/2
|
||
|
❤
|
PEMBAHASAN:
|
f(x) = x/|x|
|
(a)
|
f(0) does not
exist.
|
||
|
(b)
|
lim
x → 0
|
f(x) does
not exist.
|
|
|
(c)
|
lim
x → 0–
|
f(x) = –1
|
|
|
(d)
|
lim
x → 1/2
|
f(x) = 1
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
x → 1+
|
x2 – 1
|
does not exist.
|
|
|
|x
– 1|
|
|||
|
|
|||
|
lim
x → 1+
|
x2 – 1
|
= –2
|
|
|
|x
– 1|
|
|||
|
and
|
|||
|
lim
x → 1+
|
x2 – 1
|
= 2
|
|
|
|x
– 1|
|
|||
|
❤
|
PEMBAHASAN:
|
|
lim
x → 0
|
√(x
+ 2) – √2
|
=
|
lim
x → 0
|
[√(x
+ 2) – √2][√(x + 2) + √2]
|
|||||
|
x
|
x
[√(x
+ 2) + √2]
|
||||||||
|
|
|
=
|
lim
x → 0
|
x + 2 – 2
|
|||||
|
|
x
[√(x
+ 2) + √2]
|
||||||||
|
|
|
=
|
lim
x → 0
|
x
|
|||||
|
|
x
[√(x
+ 2) + √2]
|
||||||||
|
|
|
=
|
lim
x → 0
|
1
|
|||||
|
|
√(x
+ 2) + √2
|
||||||||
|
|
|
=
|
1
|
||||||
|
|
√(0
+ 2) + √2
|
||||||||
|
|
|
=
|
1
|
||||||
|
|
2√2
|
||||||||
|
|
|
=
|
√2
|
||||||
|
|
4
|
||||||||
|
f(x)
|
=
|
{
|
x
|
if
|
x is rational
|
|
–x
|
if
|
x is irrational
|
Find each value, if possible.
|
(a)
|
lim
|
f(x)
|
|
x → 1
|
|
|
|
(b)
|
lim
|
f(x)
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
x → 1
|
f(x) does
not exist.
|
|
|
(b)
|
lim
x → 0
|
f(x) = 0
|
|
|
BACA JUGA:
|
|
|
(a)
|
Its domain is
the interval [0, 4].
|
|||
|
(b)
|
f(0) = f(1)
= f(2) = f(3) = f(4) = 1
|
|||
|
(c)
|
lim
|
f(x) = 2
|
||
|
x → 1
|
|
|||
|
(d)
|
lim
|
f(x) = 1
|
||
|
x → 2
|
|
|||
|
(e)
|
lim
|
f(x) = 2
|
||
|
x → 3–
|
|
|||
|
(f)
|
lim
|
f(x) = 1
|
||
|
x → 3+
|
|
|||
|
❤
|
PEMBAHASAN:
|
|
f(x)
|
=
|
{
|
x2
|
if
|
x is rational
|
|
x4
|
if
|
x is irrational
|
For what value of a does limx → a f(x) exist?
|
❤
|
PEMBAHASAN:
|
|
lim
x → a
|
f(x) exists
for a = –1, 0, 1.
|
|
❤
|
PEMBAHASAN:
|
The changed values will not change limx → a f(x) at any a. As x approaches a, the limit is still a2.
|
(a)
|
lim
x → 1
|
|x
– 1|
|
||||||
|
x – 1
|
||||||||
|
(b)
|
lim
x → 1–
|
|x
– 1|
|
||||||
|
x – 1
|
||||||||
|
(c)
|
lim
x → 1–
|
x2 – |x –
1| – 1
|
||||||
|
|x
– 1|
|
||||||||
|
(d)
|
lim
x → 1–
|
[
|
1
|
–
|
1
|
]
|
||
|
x – 1
|
|x
– 1|
|
|||||||
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
x → 1
|
|x
– 1|
|
does not exist.
|
|||||||||
|
x – 1
|
||||||||||||
|
|
lim
x → 1–
|
|x
– 1|
|
= –1 and
|
|||||||||
|
x – 1
|
||||||||||||
|
|
lim
x → 1+
|
|x
– 1|
|
= 1
|
|||||||||
|
x – 1
|
||||||||||||
|
(b)
|
lim
x → 1–
|
|x
– 1|
|
= –1
|
|||||||||
|
x – 1
|
||||||||||||
|
(c)
|
lim
x → 1–
|
x2 – |x –
1| – 1
|
= –3
|
|||||||||
|
|x
– 1|
|
||||||||||||
|
(d)
|
lim
x → 1–
|
[
|
1
|
–
|
1
|
]
|
does not exist.
|
|||||
|
x – 1
|
|x
– 1|
|
|||||||||||
|
(a)
|
lim
|
√(x –〚x〛)
|
|
x → 1+
|
|
|
|
(b)
|
lim
|
〚1/x〛
|
|
x → 0+
|
|
|
|
(c)
|
lim
|
x(–1)〚1/x〛
|
|
x → 0+
|
|
|
|
(d)
|
lim
|
〚x〛(–1)〚1/x〛
|
|
x → 0+
|
|
|
❤
|
PEMBAHASAN:
|
|
(a)
|
Lim
|
√(x –〚x〛) = 0
|
|
x → 1+
|
|
|
|
(b)
|
Lim
|
〚1/x〛does not exist.
|
|
x → 0+
|
|
|
|
(c)
|
lim
|
x(–1)〚1/x〛= 0
|
|
x → 0+
|
|
|
|
(d)
|
lim
|
〚x〛(–1)〚1/x〛= 0
|
|
x → 0+
|
|
|
(a)
|
lim
|
x〚1/x〛
|
|
x → 0+
|
|
|
|
(b)
|
lim
|
x2〚1/x〛
|
|
x → 0+
|
|
|
|
(c)
|
lim
|
(〚x〛+ 〚–x〛)
|
|
x → 3–
|
|
|
|
(d)
|
lim
|
(〚x〛+ 〚–x〛)
|
|
x → 3+
|
|
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
|
x〚1/x〛= 1
|
|
x → 0+
|
|
|
|
(b)
|
lim
|
x2〚1/x〛= 0
|
|
x → 0+
|
|
|
|
(c)
|
lim
|
(〚x〛+ 〚–x〛) = –1
|
|
x → 3–
|
|
|
|
(d)
|
lim
|
(〚x〛+ 〚–x〛) = –1
|
|
x → 3+
|
|
|
(a)
|
lim
|
〚x〛/
x
|
|
x → 3
|
|
|
|
(b)
|
lim
|
〚x〛/
x
|
|
x → 0+
|
|
|
|
(c)
|
lim
|
〚x〛
|
|
x → 1.8
|
|
|
|
(d)
|
lim
|
〚x〛/
x
|
|
x → 1.8
|
|
|
❤
|
PEMBAHASAN:
|
|
(a)
|
lim
|
〚x〛/
x
: Does not exist.
|
|
x → 3
|
|
|
|
(b)
|
lim
|
〚x〛/
x
= 0
|
|
x → 0+
|
|
|
|
(c)
|
lim
|
〚x〛= 1
|
|
x → 1.8
|
|
|
|
(d)
|
lim
|
〚x〛/
x
= 0.556
|
|
x → 1.8
|
|
Many software packages have programs for calculating lim-ts, although you should be warned that they are not infallible. To develop confidence in your program, use it to recalculate some of he limits in Problems 1–28. Then for each of the following, find he limit or state that it does not exist.
|
❤
|
PEMBAHASAN:
|
limx → 0 √x does not exist since √x is not defined for x < 0.
|
❤
|
PEMBAHASAN:
|
|
lim
|
xx
|
= 1
|
|
x → 0+
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
|
√|x|
|
= 0
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
|
|x|x
|
= 1
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
|
(sin 2x)
/ 4x
|
= ½
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
|
(sin 5x)
/ 3x
|
= 5/3
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
|
cos (1 / x)
: does not exist.
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
|
x cos (1
/ x)
|
= 0
|
|
x → 0
|
|
|
❤
|
PEMBAHASAN:
|
|
lim
x → 1
|
x3 – 1
|
= 6
|
|
√(2x
+ 2) – 2
|
|
❤
|
PEMBAHASAN:
|
|
lim
x → 0
|
x sin 2x
|
= 2
|
|
sin
(x2)
|
|
❤
|
PEMBAHASAN:
|
|
lim
x → 2–
|
x2 – x – 2
|
= –3
|
|
|x
– 2|
|
|
❤
|
PEMBAHASAN:
|
|
lim
x → 1+
|
2
|
= 0
|
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1
+ 21/(x – 1)
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❤
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PEMBAHASAN:
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limx → 0 √x; The computer gives a value of 0, but limx → 0– √x does not exist.
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BACA JUGA:
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Demikian soal serta penjelasan untuk Pembahasan Soal Buku Calculus 9th Purcell Chapter 1 - 1.1 Introduction to Limit - Number 29 – 58. Silahkan untuk berkunjung kembali dikarenakan akan selalu ada update terbaru 😊😄🙏. Silahkan juga untuk memilih dan mendiskusikan di tempat postingan ini di kolom komentar ya supaya semakin bagus diskusi pada setiap postingan. Diperbolehkan request di kolom komentar pada postingan ini tentang rangkuman atau catatan atau soal dan yang lain atau bagian hal yang lainnya, yang sekiranya belum ada di website ini. Terima kasih banyak sebelumnya 👍. Semoga bermanfaat dan berkah untuk kita semua. Amiiinnn 👐👐👐
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