L’ Hospital Rule
[Pages:3]L' Hospital Rule
x 2 sin 1
1. (a) f(x) = lim
x , g(x) = sin x
x0 sin x
lim
f
'(x)
=
lim
2x sin
1 x
-
cos
1 x
x0 g'(x) x0
cos x
Since
1
limcos does not exist, we cannot apply L'hospital rule.
x0
x
(b) f(x) = x ? sin x, g(x) = x + sin x
lim f '(x) = lim 1 - cos x x g'(x) x 1 + cos x
Since
limcosx does not exist, we cannot apply L'hospital rule.
x
(c) f(x) = 2x + sin 2x, g(x) = (2x sin x) esin x lim f '(x) = lim
1 + cos2x
x g'(x) x esin x (x sin x cos x + sin x + x cos x)
Since
lim(1 + cos2x) does not exist, we cannot apply L'hospital rule.
x
1 - tan x 2. (a) lim
x cos 2x 4
( ) (0 / 0 form) = (LHR )
lim
x 4
- sec2 x - 2sin 2x
=
sec 2 2 sin
4
=
2
2 =1 2(1)
4
1 - 2 sin x (b) lim
x cos 3x 6
(0 / 0 form) =
( LHR )
lim
x 6
- 2 cos x - 3sin 3x
=
2 cos
6 3 sin
=
2
3 2
3(1)
=
3 3
2
(c)
lim
x 2
tan x tan 3x
( / form) =
( LHR )
lim
x 2
sec2 3sec2
x 3x
=
1 3
lim
x 2
cos 3x cos x
2
(
=
LHR
)
1 3
lim
x 2
- 3sin 3x - sin x
2
=
3
(d) lim( - x)tan x = lim y tan - y = lim y cot y = lim y = lim 1 = 2
x
2 y0
2
y0
2 y0 tan y ( LHR ) y0 1 sec2 y
2
22
(e)
L
=
lxim0
1 x
-
1 sin
x
3
+
3
x2
1 sin
x
-
x
1 sin2
x
=
lim x0
sin x - x x sin x
3
+
3lim x0
2sin x x2 - x2
- 2x cos2x
=
(L1
)3
+
3L2
(correspondingly)
Apply LHR to both L1 and L2 ,
L1 = lim 1 - cos x = lim
sin x
= 0 =0
(LHR ) x0 sin x + x cos x (LHR ) x0 2 cos x - x cos x 2(1) - 0
L2 =
2sin x - 2x lim
= lim
cosx - 1
x0 x 2 - x 2 cos2x (LHR) x0 x - x cos2x + x 2 sin2 x
= lim
- sin x
(LHR) x0 1 - cos 2x + 2x sin 2x + 2x sin 2x + 2x 2 cos 2x
= lim
- cosx
= -1=
(LHR) x0 2sin 2x + 4sin 2x + 8x cos2x + 4x cos2x - 4x 2 sin 2x 0
L=
1
x + sin x x - sin x
2. (f) L = lim cos(sin x) - cos x = lim 2 sin 2 sin 2
x0
x4
x0
x4
x + sin x
x - sin x
=
1 2
lim
x0
x2
- sin2 x4
x
sin lim x0 x
+
2 sin x
sin
lim
x0
2 x - sin x
=
1 2
L1L2 L3
(correspondingly)
2
2
L2 = L3 = 1 as lim sin = 1 0
L1
= lim 2x - 2sin x cosx = lim 2x - sin 2x = lim 2 - 2cos2x = lim 4sin 2x
( LHR ) x0
4x3
x0
4x3
( LHR ) x0 12x 2
( LHR ) x0 24x
= lim 8cos2x = 8 = 1
24 ( LHR ) x0
24 3
L = 1 ? 1 ?1?1= 1
23
6
(g)
x - x cosx lim
=
1 - cos x + x sin x lim
= lim 2sin x + x cosx = lim 2 +
x
cosx = 3
x0 x - sin x ( LHR ) x0
1 - cos x
( LHR ) x0
sin x
x0 sin x
(h)
tan x - x lim
=
lim
sec2
x
-1
=
lim
1 cos2
x
-1
=
1+ lim
cos x
=
1+1
=
2
x0 x - sin x (LHR) x0 1 - cos x x0 1 - cos x x0 cos2 x
1
3.
(a)
1 - cos x lim
=
lim sin x = 1
x 2x x0
2 ( LHR ) x0
2
(b)
lim
x0
x
- tan x x3
=
1 - sec2 lim
3x ( LHR ) x0
2
x
=
- lim x0
tan2 x 3x 2
=
-
1 lim 3 x0
tan x x
2
=
-
1 3
(c)
lim(sec x x
-
tan x)
=
lim x
1 cos x
-
sin x cos x
=
1 - sin x
lim
x
cos x
= lim
( LHR ) x
- cos x - sin x
=
lim cot x x
=
0
2
2
2
2
2
( ) (d)
lim
x1
x2
-1
x tan
2
=
lim
x1
x2
-1 x
cot
= lim 2x ( LHR ) x1 - csc2
x
= - lim 2x sin2 x1
x = - 2 2
2
2
(e)
lim
4+x -
4-x
=
2
1+ 4+x 2
1 4-x
=1
x0
x
( LHR )
1
2
(f)
lim
3x - 12 - x
3+ 1
1+1
= lim 2 3x 2 12 - x = 2 6 = 8
x3 2x - 3 19 - 5x ( LHR ) x3 2 +
15
2 + 15 69
2 19 - 5x
4
(g)
lim x 2 x
-
x 1
tan x
=
lyim0
1 y2
-
y
1 tan
y
=
lyim0
1 y2
-
cos y ysin y
=
lyim0
sin
y y
- ycos 2 sin y
y
= lim cos y - cosy + ysin y = lim
sin y
=
cos y
=1
(LHR) y0 y2 cos y + 2ysin y y0 y cos y + 2sin y (LHR) cos y - ysin y + 2cos y 3
Result follows by replacing x by positive integral of n .
2
4. (a) lim x + tan x = lim 1 + sec2 x = 1 + 1 = 1 x 0 sin 2x (LHR) x 0 2 cos2x 2(1)
(b) lim tan-1 x = lim y = 1
x0 x
y0 tan y
(c)
e2x -1 lim
= lim
2e2x
= 2 =2
x 0 ln (1 + x) (LHR) x 0 1/(1 + x) 1/1
(d)
ax -1 lim
=
lim a x ln a = ln a
x 1 x 0
( LHR ) x 0
(e) lim x3e-x = 3x2 = lim 6x = lim 6 = 0
x
e e e ( LHR ) x ( LHR ) x x ( LHR ) x x
(f) lim sin x ln x = lim ln x = lim 1/ x = - lim sin x tan x = - lim (sin x sec2 x + cos x tan x) = 0
x 0+
x 0+ csc x (LHR) x 0+ - csc x cot x
x 0+
x
( LHR ) x 0+
(g) lim 1 - csc x = lim 1 - 1 , see 2(e)
x0 x
x0 x sin x
(h)
lim x 1
x x -1
-
1 ln x
=
lxim1
x ln (x
x -
- 1)
x +1 ln x
(
=
LHR
)
lim
x 1
x(1/ x) + ln x
(x - 1)(1/ x) +
-1 ln x
=
lim
x 1
x
x ln x - 1 + x ln x
= lim
x(1/ x) + ln x
=
1+ lim
ln x
=
1
(LHR) x 1 1 + x(1/ x) + ln x x1 2 + ln x 2
(i)
lim (tan5x
x
-
tan x)
=
lim
x
sin 5x cos5x
-
sin x cos x
=
lim
x
sin5x cosx - cos5x sin x cos5x cosx
=
lim
x
sin 4x cos5x cosx
2
2
2
2
= lim
4 cos 4x
= 4(1) =
(LHR) x - cos5x sin x - 5sin 5x cos x - 0 - 0
2
(j)
ln (ex + x 2 ) lim
=
1 (ex + 2x)
lim ex + x 2
= lim
ex + 2x
= lim
ex + 2
x
x2
( LHR ) x
2x
x 2x(ex + x 2 ) (LHR) x 2(ex + x 2 ) + 2x(ex + 2x)
= 1 lim ex + 2
= 1 lim
ex
= 1 lim
ex
2 x ex + 3x 2 + xex 2 e (LHR) x x + 6x + ex + xex 2 x 2ex + 6x + xex
=
1 lim
ex
= 1 lim
ex
=
1 lim
ex
= 1 lim 1 = 0
(LHR) 2 x 2ex + 6 + ex + xex 2 x 3ex + 6 + xex (LHR) 2 x 3ex + ex + xex 2 x 4 + x
1
(k) y = lim x x x
ln y = lim ln x
=
1/ x lim
= lim 1
=0
y = e0
=1
x 1 x
( LHR ) x
x x
(l) y = lim xsinx x0
ln y = lim ln x = lim ln x = lim 1/ x = - lim sin x tan x x 0 1/ sin x x 0 csc x (LHR) x 0 - csc x cot x x 0 x
= - lim sin x sec2 x + tan x cosx = 0
( LHR ) x 0
1
y = e0 =1
(m) y = lim x x x 0+
1
(n) y = lim x 1-x x 1
ln y = lim x ln x = lim ln x = lim 1/ x = - lim x
x 0+
x 0+ 1/ x ( LHR ) x 0+ - 1/ x 2
x 0+
y = e0 =1
ln y = lim ln x = lim 1/ x = -lim 1 = -1
x 1 1 - x ( LHR ) x 1 - 1
x1 x
y = e-1 = 1 e
3
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- cos x cos x cos x x cos x x
- math class home
- l hospital rule
- phƢƠng trÌnh lƢỢng giÁc cƠ bẢn weebly
- trigonometria disequazioni trigonometriche
- tÀi liỆu hỌc tẬp miỄn phÍ 0 1 3 2 4
- practice problems trig integrals solutions
- integration using trig identities or a trig substitution
- with suhaag sir
- forty fourth annual