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Calculus and Analytic Geometry, MATHEMATICS 12

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Important Limits

I. 1limn n

n

x a

x ana

x a

, where n is integer and 0.a

II. 0

1lim

2x

x a a

x a

.

III. 0

1lim 1

n

ne

n

.

IV. 1

lim 1 xx

x e

.

V. 0

1lim ln

x

x

aa

x

, where 0a .

VI. 0

1lim ln 1

x

x

ee

x

.

VII. If is measured in radian, then 0

sinlim 1

.

Question # 1

(i) 3

lim(2 4)x

x

3 3

lim(2 ) lim(4)x x

x

3

2lim( ) 4x

x

2(3) 4 10 .

(ii) 21

lim 3 2 4x

x x

23(1) 2(1) 4 3 2 4 5 .

(iii) 23

lim 4x

x x

2(3) (3) 4 9 3 4 16 4 .

(iv) 22

lim 4x

x x

22 2 4 = 0.

(v) 3 22

lim 1 5x

x x

3 22 2

lim 1 lim 5x x

x x

3 2(2) 1 (2) 5 8 1 4 5 9 9 0 .

(vi) 3

2

2 5lim

3 2x

x x

x

32( 2) 5( 2)

3( 2) 2

16 10

6 2

26

8

13

4 .

Question # 2

(i) 3

1lim

1x

x x

x

2

1

( 1)lim

1x

x x

x

1

( 1)( 1)lim

1x

x x x

x

1

lim ( 1)x

x x

( 1)( 1 1) 2

(ii) 3

20

3 4limx

x x

x x

=

2

0

(3 4)lim

( 1)x

x x

x x

2

0

3 4lim

1x

x

x

3(0) 44

0 1

.

FSc-II / Ex- 1.3 - 2

(iii) 3

22

8lim

6x

x

x x

3 3

22

(2)lim

3 2 6x

x

x x x

2

2

( 2)( 2 4)lim

( 3) 2( 3)x

x x x

x x x

2

2

( 2)( 2 4)lim

( 3)( 2)x

x x x

x x

2

2

( 2 4)lim

( 3)x

x x

x

2(2) 2(2) 4)

(2 3)

12

5

(iv) 3 2

31

3 3 1limx

x x x

x x

3

21

1lim

( 1)x

x

x x

3

1

1lim

( 1)( 1)x

x

x x x

2

1

1lim

( 1)x

x

x x

2

1

1 1lim

(1)(1 1)x

0

(v) 3 2

21lim

1x

x x

x

2

1

1lim

( 1)( 1)x

x x

x x

2

1lim

( 1)x

x

x

21

( 1 1)

1

2

(vi) 2

3 24

2 32lim

4x

x

x x

2

24

2( 16)lim

( 4)x

x

x x

24

2( 4) 4lim

( 4)x

x x

x x

24

2( 4)limx

x

x

2

2(4 4)

4

161

16 .

(vii) 2

2lim

2x

x

x

2

2 2lim

2 2x

x x

x x

2 2

2

2lim

2 2x

x

x x

22

lim2 2x

x

x x

2

1lim

2x x

1

2 2

1

2 2

(viii) 0

limh

x h x

h

0limh

x h x x h x

h x h x

2 2

0limh

x h x

h x h x

0limh

x h x

h x h x

FSc-II / Ex- 1.3 - 3

0

limh

h

h x h x

0

1limh x h x

1

0x x

1

2 x

(ix) limn n

m mx a

x a

x a

1 2 3 2 1

1 2 3 2 1

....lim

....

n n n n

m m m mx a

x a x x a x a a

x a x x a x a a

1 2 3 2 1

1 2 3 2 1

....lim

....

n n n n

m m m mx a

x x a x a a

x x a x a a

1 2 3 2 1

1 2 3 2 1

....

....

n n n n

m m m m

a a a a a a

a a a a a a

1 1 1 1

1 1 1 1

.... ( terms)

.... ( terms)

n n n n

m m m m

a a a a n

a a a a m

1

1

n

m

na

ma

1 1n m

na

m

n mn

am

Law of Sine

If is measured in radian, then 0

sinlim 1

See proof on book at page 25

Question # 3

(i) 0

sin7limx

x

x

Put 7t x 7

tx

When 0x then 0t , so

0 0

sin7 sinlim lim

7x t

x t

tx

0

sin7lim

t

t

t 7(1) 7 By law of sine.

(ii) 0

sinlimx

x

x

Since 180 rad 1180

rad

180

xx

rad

So 0

sinlimx

x

x

0

sin180lim

x

x

x

FSc-II / Ex- 1.3 - 4

Now put 180

xt

i.e.

180tx

When 0x then 0t , so

0 0

sin sin180lim lim180x x

xt

tx

0

sinlim

180 x

t

t

(1)180

180

by law of sine

(iii) 0

1 coslim

sin

0

1 cos 1 coslim

sin 1 cos

2

0

1 coslim

sin 1 cos

2

0

sinlim

sin 1 cos

0

sinlim

1 cos

sin(0)

1 cos(0)

0

1 1

0

(iv) sin

limx

x

x

Put t x x t

When x then 0t , so

0

sin sin( )lim limx t

x t

x t

0

sinlimt

t

t sin sin 2

2t t

sin t

1 By law of sine.

(v) 0

sinlim

sinx

ax

bx

0

1limsin

sinxax

bx

0

1limsin

sinx

axax

bxaxbx

bx

0

sin 1lim

sinx

axax

bxaxbx

bx

0

0

sin 1lim

sinlim

x

x

a ax

bxb ax

bx

1

(1)(1)

a

b

a

b by law of sine

(vi) 0

limtanx

x

x 0lim

sin

cos

x

x

x

x

0lim cos

sinx

xx

x

0

1lim cos

sinxx

x

x

0

0

1limcos

sinlim

x

x

xx

x

1

1 11

(vii) 20

1 cos2limx

x

x

2

20

2sinlimx

x

x

2 1 cos 2sin

2

xx

22sin 1 cos2x x

FSc-II / Ex- 1.3 - 5

2

0

sin2lim

x

x

x

2

0

sin2 lim

x

x

x

22(1) 2

(vii) Do yourself by rationalizing

(viii) 2

0

sinlim

0

sinlim sin

0 0

sinlim limsin

(1) (0) 0

(x) 0

sec coslimx

x x

x

0

1cos

coslimx

xx

x

2

0

1 cos

coslimx

x

x

x

2

0

1 coslim

cosx

x

x x

2

0

sinlim

cosx

x

x x

0

sin sinlim

cosx

x x

x x

0 0

sin sinlim lim

cosx x

x x

x x

sin(0)1

cos(0)

01

1 0

(xi) 0

1 coslim

1 cosx

p

q

2

0 2

2sin2lim

2sin2

x

p

q

2

1 cossin

2 2

x x

20 2

1lim sin

2sin

2

x

p

q

2

2

2 20

2

2

2

2 2

2

1lim sin

2

sin .2

x

p

p q

q

p

q

22

20 2

2

2

2

2

2

2

sin12lim

sin2 .

x

p

p

q

q

p

q

22 2

2 2 20

sin12 4lim

sin 42 2

2

x

p p

p qq

q

2

2

22 0

0

sin12lim

sin2 2lim

2

x

x

pp

pq q

q

2

2

2 2

1(1)

(1)

p

q

2

2

p

q

FSc-II / Ex- 1.3 - 6

(xii) 30

tan sinlim

sin

30

sinsin

coslimsin

30

sin sin cos

coslimsin

30

sin sin coslim

sin cos

30

sin 1 coslim

sin cos

20

1 coslim

sin cos

20

1 cos 1 coslim

sin cos 1 cos

2

20

1 coslim

sin cos 1 cos

2

20

sinlim

sin cos 1 cos

0

1lim

cos 1 cos

01

limcos 1 cosx

1

cos(1) 1 cos(1)

1

1 1 1

1

2

Note:

a) 1

lim 1

n

ne

n

b) 1

0lim 1 xx

x e

where 2.718281...e

See proof of (a) and (b) on book at page 23

c) 0

1lim log or ln

x

ex

aa a

x

Proof:

Put 1xy a .. (i)

When 0x then 0y

Also from (i) 1 xy a

Taking log on both sides

ln 1 ln xy a ln(1 ) lny x a ln lnmx m x

ln 1ln

yx

a

Now 0 0

1lim lim

ln 1

ln

x

x y

a y

yx

a

0ln

limln 1y

y a

y

0

lnlim

1ln 1

y

a

yy

FSc-II / Ex- 1.3 - 7

1

0

lnlim

ln 1y

y

a

y

1

0

ln

limln 1 yy

a

y

ln lnmx m x

1

0

ln

ln lim 1 yy

a

y

ln

ln

a

e

1

0lim 1 xx

x e

ln

1

a ln a ln 1e

Question # 4

(i)

21

lim 1

n

n n

2

1lim 1

n

n n

2e

(ii) 21

lim 1

n

n n

1

21lim 1

n

n n

1

2e e

(iii) 1

lim 1

n

n n

1

1lim 1

n

n n

1e 1

e

(iv) 1

lim 13

n

n n

3

31lim 1

3

n

n n

13 31

lim 13

n

n n

1

3e

(v) 4

lim 1

n

n n

44

4 44 4lim 1 lim 1

n n

n nn n

4e .

(vi) 2

0lim 1 3 xx

x

6

3

0lim 1 3 xx

x

6

1

3

0lim 1 3 xx

x

6e

(vii) 21

2

0lim 1 2 xx

x

22

2 2

0lim 1 2 xx

x

22

12 2

0lim 1 2 xx

x

2e

(viii) 1

0lim 1 2 hh

h

2

2

0lim 1 2 hh

h

21

2

0lim 1 2 hh

h

2e 2

1

e

(ix) lim1

x

x

x

x

1

lim

x

x

x

x

1lim

x

x

x

x x

1lim 1

x

x x

FSc-II / Ex- 1.3 - 8

1

1lim 1

x

x x

1e1

e

(x)

1

10

1lim

1

x

x x

e

e

; 0x

Put x t where 0t When 0x then 0t , so

1

10

1lim

1

x

x x

e

e

1

10

1lim

1

t

t t

e

e

10

10

1

1

e

e

1

1

e

e

0 1

0 1

1 10e

e

1

(xi)

1

10

1lim

1

x

x x

e

e

; 0x

1

1

0 1

1

11

= lim1

1

x

x

xx

x

ee

ee

1

0

1

11

= lim1

1

x

x

x

e

e

1

0

10

11

=1

1

e

e

11

=1

1

e

e

11

=1

1

1 0

=1 0

1

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Book: Exercise 1.3 (Page 27)

Calculus and Analytic Geometry Mathematic 12

Punjab Textbook Board, Lahore.

Edition: May 2013.

Made by: Atiq ur Rehman (Atiq@MathCity.org)

Available online at http://www.MathCity.org in PDF Format (Picture format to view

online).

Page Setup used A4.

Printed: December 19, 2014.