(15) 5.00 _d²_ dx x²+ 100 = 5.00 _d² + dx - 2 X +100 = lim 848 = lim 848 b = lim Sa dx + lim so dx 948 x² + 10² b→∞ + [ lo So di dx = 1 b So de + lim so de dx a x² +100 640 x + 100 = 1 x + 100 I tam' x 10 -] 10 | a [ +++] 10 2 2 + lim I tan! •÷ [(*)-(+)] lim tan x 'lim tam' x 10 10 a b→∞0 10 o- lim tom' a + lim tam' b a-8 b→∞ 10 10 O - O s dx = ____ tan't a Lom trying to understand why tań" (00) = I/I
(15) 5.00 _d²_ dx x²+ 100 = 5.00 _d² + dx - 2 X +100 = lim 848 = lim 848 b = lim Sa dx + lim so dx 948 x² + 10² b→∞ + [ lo So di dx = 1 b So de + lim so de dx a x² +100 640 x + 100 = 1 x + 100 I tam' x 10 -] 10 | a [ +++] 10 2 2 + lim I tan! •÷ [(*)-(+)] lim tan x 'lim tam' x 10 10 a b→∞0 10 o- lim tom' a + lim tam' b a-8 b→∞ 10 10 O - O s dx = ____ tan't a Lom trying to understand why tań" (00) = I/I
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I am trying to understand why is tan^-1(∞) = π/2
![(015) 5.00
:
5.00 dr
dx
2
x + 100
10
2
X + 100
lim
a →∞0 10
= 1
b
= lim Sa dx + lim So do
dx
a →∞
X + 100
b
+ lim so
b→∞
10
+
= lim So dx + lim
a48
x² +10²
*+[..
lo
00
5.⁰⁰
O
dx
2
x + 100
2
x + 100 640
[[|]
+ lim
b→∞0 10
tam x
I + I
2
10 | a
dis
dx
x + 10
tan_
[(²)(²+1)]
'lim tan x
lim
10 a
tan
。- lim tam' a + lim tan b
b→∞
a →∞0
10
O
b
10 O
S do = I tom' I
dx
x² + a²
a
a
Lom trying to understand why tań" (co) = III
am
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3073de58-1b40-421a-8538-3fd557e8e327%2Fa6b1f59f-bbd6-468d-8196-5275f69c95ce%2F689q1tt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(015) 5.00
:
5.00 dr
dx
2
x + 100
10
2
X + 100
lim
a →∞0 10
= 1
b
= lim Sa dx + lim So do
dx
a →∞
X + 100
b
+ lim so
b→∞
10
+
= lim So dx + lim
a48
x² +10²
*+[..
lo
00
5.⁰⁰
O
dx
2
x + 100
2
x + 100 640
[[|]
+ lim
b→∞0 10
tam x
I + I
2
10 | a
dis
dx
x + 10
tan_
[(²)(²+1)]
'lim tan x
lim
10 a
tan
。- lim tam' a + lim tan b
b→∞
a →∞0
10
O
b
10 O
S do = I tom' I
dx
x² + a²
a
a
Lom trying to understand why tań" (co) = III
am
2
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