L’ Hospital Rule

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

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