LIMITS PART 1 of 1

18 Pages·2016·0.36 MB·English

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STUDY PACKAGE
Subject : Mathematics
Topic : LIMITS
Available Online : www.MathsBySuhag.com
R
Index
1. Theory
2. Short Revision
3. Exercise (Ex. 1 + 5 = 6)
4. Assertion & Reason
5. Que. from Compt. Exams
6. 39 Yrs. Que. from IIT-JEE(Advanced)
7. 15 Yrs. Que. from AIEEE (JEE Main)
Student’s Name :______________________
Class :______________________
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Limit
1. Limit of a function f(x) is said to exist as,
x  a when,
m 8
o Lhim0it f (a  h) = Lhim0it f (a + h) = some finite value M. f 1
o
c (Left hand limit) (Right hand limit)
. 2
g Note that we are not interested in knowing about what happens at x = a. Also note that if L.H.L. &e
ha R.H.L. are both tending towards '  ' or ‘–’ then it is said to be infinite limit. pag
ySuSolved RExeammepmlbee #r, 1Lximait x  a .
B 1
8
s 8
h 8
5
at 0
M 3
Limit 9
. Find f(x) 8
w x/2 9
w 0
w
,
Solution. 9
& 7
m Here Limit f(x) = 1 77
x/2
3
oSolved Example # 2 0
c 9
. 3
s
0
e 9
s 0
s
a e :
l n
C Find Limit f(x) o
o x1 h
P
k

e al
T p
.Solution. o
w h
w Left handed limit = 1 Right handed limit = 2 B

wSolved HEexnacmeplLexim #1i t3 f(x) = does not exist. Sir),
: .
e limit limit limit K
it (i) Find x0 f(x) (ii) Find x1 f(x) (iii) Find x3 f(x) R.
s

b S.
e (

w a
y
m ri
a
o K

fr R.
e g
g a
a h
u
k S
c
a s :
P h
Solution. t
a
y limit M
d (i) f(x) = does not exists
u bxec0ause left handed limit  right handed limit s,
t e
S limit limit s
(ii) f(x) = 0 (iii) f(x) = 1 s
d x1 x3 a
a2. Indeterminant Forms: Cl
o o
nl 0 ,  , 0 , º, 0º,and 1. ek
w 0  T
o
Solved Example # 4
D
Which of the following limits are forming indeterminant from also indicate the form

E
1 1x
E (i) lim (ii) lim
R x0 x x0 1x2
F 1 1 
(iii) lim x n x (iv) lim   
x0 x0 x x2 
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lim lim
(v) (sin x)x (vi) (n x)x
x0 x0
1
(vii) lim 1sinxx (viii) lim (1)1/x
x0 x0
Solution
0
(i) No (ii) Yes from
m 0 8
1
o (iii) Yes 0 ×  form (iv) Yes ( – ) form f
c (v) Yes, (0)º form (vi) Yes ()º form o
. (vii) Yes (1) form (viii) 3
g e
a NOTE : g
h (i) '0' doesn't means exact zero but represent a value approaching towards zero similary to '1' anda
BySu ((iiiii)) i nf i+ nx i ty. = =   1. p
(iv) (a/) = 0 if a is finite 8
s 8
h a 58
at (v) 0 is not defined for any a  R. 0
M 3
(vi) a b = 0, if & only if a = 0 or b = 0 and a & b are finite. 9
. 8
w
3. Method of Removing Indeterminancy 9
w 0
w
To evaluate a limit, we must always put the value where 'x' is approaching to in the function. If we get9,
& a determinate form, then that value becomes the limit oth er wise if an indeterminant form comes. Then7
m apply one of the following methods: 77
(i) Factorisation (ii) Rationalisation or double rationalisation 3
o (iii) Substitution (iv) Using standard limits 0
c 9
. (v) Expansions of functions. 3
s
0
e1. Factorization method :- 9
s We can cancel out the factors which are leading to indeterminancy and find the limit of the remaing0
s
a expression. e :
l n
C
Solved Example # 5 o
o h
Tek lxim2it xx63224xx1126 pal P
.Solution. o
w h
w x6 24x16 (x2)(x5 2x4 4x3 8x2 16x8) B
limit = limit ,
w x2 x3 2x12 x2 (x2 2x6)(x2) r)
168 Si
e: = = 12 K.
t 14 .
i R
s
b2. Rationalization /Double Rationalization. S.
e (
w We can rationalize the irrational expression by multiplying with their conjugates to remove thea
m indeterminancy. riy
a
o K
4 5x1
frSolved Example # 6 limit . R.
e x1 2 3x1 g
g a
aSolution. limit 4 5x1 uh
k x1 2 3x1 S
c
a (4 5x1)(2 3x1)(4 5x1) s :
P limit h
= t
x1 (2 3x1)(4 5x1)(2 3x1) a
y M
d
u limit (155x) 2 3x1 5 s,
St = x1 (33x) × 4 5x1 = 6 se
s
dSolved Example # 7 a
a Cl
nlo Evaluate : (i) xlim2 x12 x32(23xx23)2x (ii) xlim0 1x x 1x eko
w   T
(2x3) x 1
o (iii) lim
D x1 2x2 x3
Solution (i) We have
E
E  1 2(2x3)   1 2(2x3) 
R xlim2 x2 x3 3x2 2x = xlim2 x2 x(x1)(x2)
F
x(x1)2(2x3)
lim
=  
x2  x(x1)(x2) 
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 x2 5x6 
= lim  
x2 x(x1)(x2)
(x2)(x3)  x3  1
lim lim
=   =   = –
x2 x(x1)(x2) x2 x(x1) 2
0
(ii) The given limit taken the form when x  0. Rationalising the numerator, we get
0
m 8
.co xlim0 1x x 1x = xlim0  1x x 1x  11xx  11xx 4 of 1
g e
a  (1x)(1x)  g
h = lim    pa
ySu x0 x 1x2x 1x   2  2 .
B = lim    = lim   = = 1 81
s x0 x 1x  1x  x0  1x  1x 2 8
h (iii) We have 8
5
at (2x3) x 1 (2x3) x 1 0
.M xlim1  2x2 x3  = xlim1  (2x3)(x1)  893
w
  9
w  (2x3) x 1  0
w = lim    
x1 (2x3) x 1 x 1 9,
& 7
m = lim  2x3  = 1 = 1 77
o x1 (2x3) x 1 (5)(2) 10 03
c 9
s.4. Fundamental Theorems on Limits: 3
0
e 9
s LetLimit f (x) = &Limit g (x) = m. If & m exists then: 0
s xa xa
a Limit Limit e :
l (i) { f (x) ± g (x) } = ± m (ii) { f(x). g(x) } = . m n
C xa xa
o
o f (x)  h
k (iii) Limit = , provided m  0 P
e xa g(x) m al
T p
. (iv) Limit k f(x) = kLimit f(x) ; where k is a constant. o
w xa xa h
w   B
w (v) Limit f [g(x)] = f Limitg(x) = f (m); provided f is continuous at g (x) = m. r),
xa  xa  Si
: Solved Example # 8 Evaluate .
e K
sit (i) lim (x + 2) (ii) lim x(x – 1) (iii) lim x2 4 (iv) lim cos (sin x) R.
b x2 x2 x2 x2 x0 S.
we (v) lim x2 3x2 (vi) lim x2 3x2 a (
m x1 x2 1 x1 x2 1 riy
lim a
oSolution (i) x + 2 being a polynomial in x, its limit as x  2 is given by x2 (x + 2) = 2 + 2 = 4 K
fr (ii) Again x(x – 1) being a polynomial in x, its limit as x  2 is given by R.
e lim g
x(x – 1) = 2 (2 – 1) = 2
g x2 a
a h
x2 4 (2)2 4 u
k (iii) By (II) above, we have lim = = 2 S
ac x2 x2 22 s :
P (iv) lim cos (sin x) = cos limsinx = cos 0 = 1 th
y x0 x0  Ma
d (v) Note that for x = 1 both the numerator and the denominator of the given fraction vanish. Therefore

tu lim x2 3x2 lim (x1)(x2) lim x2 1 es,
S by (III) above, we have = = = – s
x1 x2 1 x1 (x1)(x1) x1 x1 2 s
d (vi) Note that for x = 1, the numerator of the given expression is a non-zero constant 6 and thea
a Cl
o 6 o
l denominator is zero. Therefore, the given limit is of the form . Hence, by (IV) above, wek
n 0 e
w T
x2 3x2
o conclude that lim does not exist
D x1 x2 1
E 5. Standard Limits: Limit sinx Limit tanx Limit tan1x Limit sin1x
(a) = 1 = = =
E x0 x x0 x x0 x x0 x
R [ Where x is measured in radians ]
F
x
 1
(b) Limit (1 + x)1/x = e ; Limit 1  = e
x0 x  x
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Limit ex 1 Limit ax 1
(c) = 1; = log a, a > 0
x0 x x0 x e
n(1x) xn an
Limit Limit
(d) = 1 (e) = nan – 1.
x0 x xa xa
sin2x
Limit
Solved Example # 9: Find
x0 x
m sin2x sin2x 8
Solution. Limit  Limit . 2 = 2 1
o x0 x x0 2x f
o
g.cSolved Example # 10:Limit e3x 1 e 5
a x0 x/2 g
h a
BySuSolution. Lxim0it ex3Lxi/m2i1t tanxsinLxxim0it 2 × 3 e33xx1 = – 6. 1. p
Solved Example # 11 8
s x0 x3 8
h 8
tanxsinx 5
MatSolution. Lxim0it x3 30
9
w. = Limit tanx(1cosx) 98
w x0 x3 0
w  x 2
om & = Lxim0it tanx.x23sin2 2x = Lxim0it tanxx . sin2x2 = 1. 03 7779,
c sin2x 9
.Solved Example # 12 Compute lim 3
s x0 sin3x 0
e 9
s sin2x sin2x 2x 3x  0
asSolution We have xlim0 sin3x = xlim0  2x .3x . sin3x e :
l n
C  sin2x 2  3x  o
o =  lim  . .  lim , x  0 h
k 2x0 2x  3 3x0 sin2x P

w.Te = 1 . 32 + 3lxim0 si3nx3x = 32 × 1 = 32 hopal
w x B
wSolved Example # 13 Evaluate lim 1 2 r),
x  x Si

site:Solution xlim 1 2xx = exlim2x.x = e2. R. K.

eb lim ex e3 lim x(ex 1) (S.
Solved Example # 14 Compute (i) (ii)
w x3 x3 x0 1cosx a
y
m Solution (i) Put y = x – 3. So, as x  3, y  0. Thus ri
a
o ex e3 e3y e3 K
e fr xlim3 x3 = ylim0 y g R.
g e3 .ey e3 a
a = lim h
y0 y u
k S
ac lim ey 1 s :
P = e3 y0 y = e3 . 1 = e3 th
a
y (ii) We have M
d
u lim x(ex 1) lim x(ex 1) s,
t (ii) = e
S x0 1cosx x0 x s
2sin2 s
d 2 a
a Cl
 
o ex 1 x2  o
nl 1 lim  .  ek
w = 2 . x0  x sin2 x = 2. T
o  2
D
x3 8
lim
ESolved Example # 15 Evaluate
x2 x2 4
E
Solution (First Method)
R
The given expression is of the form
F
x3 (2)3 x3 (2)3 x2 (2)2
= 
x2 (2)2 x2 x2
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x3 8 x3 (2)3 x2 (2)2
lim lim lim
 = 
x2 x2 4 x2 x2 x2 x2
xn an
lim
= 3(22)  2(21) (using = nan–1 )
xa xa
= 12  4 = 3
(Second Method)
The numerator and denominator have a common factor (x – 2). Cancelling this factor, we obtain
m 8
x3 8 x2 2x4 x3 8 x2 2x4 1
o =  lim = lim f
c x2 4 x2 x2 x2 4 x2 x2 o
g. (2)2 2(2)4 12 e 6
a = = = 3 g
h 22 4 a
Note : Since x  2, x – 2 is not zero, so the cancellation of the factor x – 2 in the above example isp
ySu6. cUasrreie do ofu t.Subsitution in Solving Limit Problems .
B 1
lim 8
s Sometimes in solving limit problem we convert f(x) by subtituting x = a + h or x = a – h as8
h xa 8
at hlim0 f(a + h) or hlim0 f(a – h) according as need of the problem. 0 5
M 3
9
w. limit 1tanx 8
Solved Example # 16 9
w x/4 1 2sinx 0
w  
Solution. Put x = + h  x   h  0 9,
& 4 4 7
m   77
1tan h
o 4  03
s.c lhim0it 1 2sin h 3 9
e 4  90
s 1tanh 0
s 1
a limit 1tanh e :
= =
l n
C h0 1sinhcosh
o
o 2tanh Ph
k

e 1tanh al
w.T = lhim0it 2sin2 h 2sinhcosh hop
w 2 2 2 B

w = lhim0it 2sin2 h22stainnhhcosh (1t1anh) Sir),
e: 2 2 2 K.
t .
i tanh R
s 2
b h 1 2 S.
limit
we = h0 sinh (1tanh) = 1 = 2. a (
m h2 sinh2cosh2 ariy
o K
2
fr7. Limit When x  R.
e g
g 1 a
a Since x     0 hence in this type of problem we express most of the part of expressionh
x u
k S
c 1 1
a in terms of and apply  0. We can see this approch in the given solve examples. s :
P x x h
t
1 a
ySolved Example # 17 limit x sin M
d x x
u s,
t limit 1 e
SSolution. x sin s
x x s
d a
a limit sin1/x Cl
= = 1
o x 1/x o
l k
n limit x2 e
wSolved Example # 18 T
x 2x3
o
D x2
limit
Solution.
E x 2x3
E 12/x 1
limit
R = .
x 23/x 2
F
x2 4x5
limit
Solved Example # 19
x 3x2 x3 2
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x2 4x5
limit
Solution.
x 3x2 x3 2
1 4 5
 
x x2 x3
limit
= 3 2 = 0
x 1
x x3
mSolved Example # 20 lim 3x2 2 18
o x x2 of
c
g.Solution. lim 3x2 2 e 7
a x x2 g
h a
BySu Put x = t132t2. 1 x  – t  0+ 81. p
s t2 8
h lim 8
at = x0 12t 0 5
M t 3
9
w. lim 32t2 t 3 98
w = x0 (12t) |t| = 1 = – 3 . 0
w8. Limits Using Expansion
,
9
& 7
m (i) ax1 xlna  x2ln2a  x3ln3a .........a0 (ii) ex 1x  x2  x3 ...... 77
1! 2! 3! 1! 2! 3!
3
o 0
c x2 x3 x4 x3 x5 x7 9
. (iii) ln (1+x) =x   .........for1x1 (iv) sinxx   ..... 3
s 2 3 4 3! 5! 7!
0
sse (v) cosx1 x2  x4  x6 ..... (vi) tan x = x x3  2x5 ...... 0 9
a 2! 4! 6! 3 15 e :
Cl x3 x5 x7 12 12.32 12.32.52 n
(vii) tan-1x = x   .... (viii) sin-1x = x x3  x5 x7..... o
o 3 5 7 3! 5! 7! h
P
k
e x2 5x4 61x6 al
T (ix) sec-1x =1   ...... p
. 2! 4! 6! o
w h
w n(n1) n(n1)(n2) B
w (x) for |x| < 1, n  R (1 + x)n = 1 + nx + 1.2 x2 + 1.2.3 x3 + ............ ),
r
te: Solved Example # 21 xlim0 ex x12x . K. Si
i R
bsSolution. lim ex 1x S.
e x0 x2 (

w a
 x2  y
m 1x 2!.......1x 1 ari
o = lim   = K
fr x0 x2 2 R.
e lim tanxsinx g
gSolved Example # 22 a
a x0 x3 h
u
k tanxsinx S
lim
cSolution.
a x0 x3 s :
P h
 x3   x3  t
dy = lim x 3 ........x 3!....... = 1 + 1 = 1. Ma
u x0 3 6 2 s,
t x3 e
S s
d Solved Example # 23 lim 7x)1/3 2 as
a x0 x1 Cl
wnloSolution. Put x  1 + h hlim0 (8h)h1/3 2 Teko
o  h1/3
D 2.1  2
E hlim0  8
h
E
R  11 h2 
  1  
F  1 h 33 8 
21 .  .......1
 3 8 1.2 
= lim  
h0
h
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1 1
lim
= 2 × =
h0 24 12
x2
n(1x)sinx
Solved Example # 24 lim 2
x0
xtanxsinx
x2
n(1x)sinx
m lim 2 8
x0
xtanxsinx 1
o f
o
c  x2 x3   x3 x5  x2
g. x 2  3 .....x 3!  5! ..... 2 e 8
a lim     1 1 1 g
ySuh9. =L ixmi0t s of formx 3.taxnx.sixnx = 3 + 6 = 2 . pa
B 1
0 8
s All these forms can be convered into form in the following ways 8
h 0 8
5
at (i) If x  1, y  , then z = (x)y 0
M nx 3
9
w.  n z = y n x  n z = (1/y) 8
9
w 1 0
w Since y   hence  0 and x  1 hence nx  0
y 9,
& (ii) If x  0, y  0, then z = xy  n z = y n x 7
7
m y 0 7
o  n z = 1/ny = 0 form 03
c 9
. (iii) If x   , y  0, then z = xy  n z = y n x 3
s
e  n z = y = 0 form 90
s 1/nx 0 0
s
a also for (1) type of problems we can use following rules. e :
Cl (i) lim (1 + x)1/x = e (ii) lim [f(x)]g(x) n
x0 xa o
o where f(x)  1 ; g(x) as x  a h
P
k
.Te = xlima 1f(x)1f(x1)1{f(x)1}.g(x) = exlima[f(x)1]g(x) opal
w 4x2 h
w  2x2 1 B
lim  
Solved Example # 25 ,
wSolution. Since it is in the xform o2fx 123 Sir)
: .
bsite xlim 22xx22314x2 = exlim 2x2 2x122x32 3 (4x2 + 2) = e–8 S. R. K
e (
w lim a
m Solved Example # 26:x4 (tan x)tan 2x ariy
o lim (tanx1)tan2x K
frSolution = ex4 R.
e 2tanx g
g lim(tanx1) a
a = ex4 1tan2x uh
k S
ac = e21(1tantan/4/4) = e–1 = 1 s :
P e h
dy Solved Example # 27 Evaluate lim 2atan2ax . Mat
u xa  x s,
t e
S x s
ad Solution. xlima 2axtan2a Clas
o o
wnl put x = a + h  lim 1 h tan22ah Tek
o h0  (ah)
D
h
E  lim 1 h cot2a  ehlim0cot2ah.1ahh1
E h0  ah
R
 h  2a
F  
lim 2a . 
h0 tanh ah
 e  2a = e–2/
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lim
Solved Example # 28: xx
x0
lim
Solution. y = xx
x0
lim
n y = x n x
x0
1
n
1
m = lim – x = 0   y = 1 8
o x0 1 x f 1
o
c x
g.10. Sandwich Theorem or Squeeze Play Theorem: e 9
a g
h If f(x)  g(x)  h(x)  x & Limit f(x) = =Limit h(x) thenLimit g(x) = . a
xa xa xa p
ySu .
B 1
8
s 8
h 8
5
at 0
M 3
9
. 8
w
9
w 0
wSolved Example # 29: Evaluate lim [x][2x][3x]....[nx] 9,
& n n2 7
m Where [ ] denotes the greatest integer function. 77
Solution. 3
o We know that, x – 1 < |x|  x 0
c  2x – 1 < [2x]  2x 9
. 3
s  3x – 1 < [3x]  3x
0
e .................... 9
s .................... 0
s
a  n(xx +– 21x < + [ n3xx] + .n..x. + nx) – n < [x] + [2x] + ..... +[nx]  (x + 2x + .... + nx) e :
l n
C
o  xn(n1) – n < n [r x]  x.n(n1) Pho
k 2 2
e r1 al
T p
w. Thus, lim [x][2x]....[nx] ho
w n n2 B

w x  1 1 [x][2x]....[nx] x  1 ),
 nlim 2 1n – n < nlim n2  nlim 2 1n Sir
: .
e x [x][2x]....[nx] x [x][2x]....[nx] x K
t  < lim   lim = .
si 2 n n2 2 n n2 2 R

bAliter We know that [x] = x – {x} S.
e n (
m w r x = [x] + [2x] + .... + nx – [nx] riya
r1 a
o = (x + 2x + 3x + ... + nx) – ({x} + {2x} + .... + {nx}) K

fr xn(n1) R.
e = 2 – ({x} + {2x} + .. + {nx}) g
g a
a n h
k  1 [r x] = x 1 1 – {x}{2x}....{nx} Su
ac n2 r1 2  n n2 s :
P n h
[r x] at
y Since, 0  {rx} < 1,  0  < n M
d r1
u n n n s,
d St  lim r1[rx] = 0  lim r1[rx] = lim x 1 1 – lim r1{rx} asse
a n n2 n n2 n 2  n n n2 Cl
o o
l n k
n [rx] e
w lim x T
o n r1 = 2
D n2
E limit 1
Solved Example # 30 x sin
E x0 x
R 1
limit
FSolution. x sin
x0 x
= 0 × (some value in [– 1, 1]) = 0
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11. Some Important Notes :
nx x
lim lim
(i) = 0 (ii) = 0
x x x ex
As x  , n x increnes much slower than any (+ve) power of x where ex increases much faster than
(+ve) power of x
(iii) Limit (1  h)n = 0 & Limit (1 + h)n  where h > 0.
n n
8
m (iv) If Limit f(x) = A > 0 & Limit (x) = B (a finite quantity) then; 1
o xa xa of
c Limit [f(x)](x) = ez where z =Limit (x). ln[f(x)] = eBlnA = AAB 0
. xa xa 1
g
e
a
g
h x1000 a
uSolved Example # 31 lim p
BySSolution. lim x10x00 = 0ex 1.
s x ex 8
8
h 8
__________________________________________________________________________________________
t 5
a Short Revesion (LIMIT) 0
M 3
w.THINGS TO REMEMBER : 89
1. Limit of a function f(x) is said to exist as, xa when 9
w
0
w Lximait f(x) = Lximait f(x) = finite quantity.. 9,
&2. FUNDAMENTAL THEOREMS ON LIMITS : 7
7
m Let Limit f (x) = l & Limit g (x) = m. If l & m exists then : 7
o(i) Limitx f (ax) ± g (x) = l ±x ma (ii) Limit f(x). g(x) = l. m 03
c xa xa 9
. f(x) 
es(iii) Lximait g(g)  m , provided m  0 903
s
s(iv) Limit k f(x) = kLimit f(x) ; where k is a constant. 0
a xa xa :
e
l
C   n
(v) Limit f [g(x)] = fLimit g(x) = f (m) ; provided f is continuous at g (x) = m. o
o xa  xa  h
k P
w.Te For example Lximait l n (f(x) = l nLximait f(x) l n l (l > 0). REMEMBER opal
Limit h
w B
 x  a
w xa ),
r
Si
: 3. STANDARD LIMITS :
e sinx tanx tan1x sin1x K.
bsit(a) [L xWim0hiterex x is = m 1e a=s uLrxeimd0i itn raxdia n=sL ]xim0it x =Lxim0it x S. R.
e
w  1x Limit a (
(b) Limit (1 + x)1/x = e = Limit 1  note however there h0 (1 - h )n = 0 y
m x0 x  x n ri
a
o Limit K
fr and hn0 (1 + h )n  R.
e(c) If Limit f(x) = 1 and Limit (x) = , then ; g
g xa xa a
a Limit h
k Limit f(x)(x)e xa (x)[f(x)1] Su
c xa

a(d) If Limit f(x) = A > 0 & Limit (x) = B (a finite quantity) then ; s :
P xa xa h
dy Lximait [f(x)](x) = ez where z = Lximait (x). ln[f(x)] = eBlnA = AAB Mat
u Limit ax 1 Limit ex1 s,
St(e) x0 x = 1n a (a > 0). In particular x0 x = 1 se
ad (f) Lximait xxnaan nan1 Clas
o
o
nl4. SQUEEZE PLAY THEOREM :If f(x)  g(x)  h(x)  x & Limit f(x) = l = Limit h(x) then Limit g(x) = l.k
w xa xa xa e
0  T
o5. INDETERMINANT FORMS : , , 0x , 0,  ,  and 1
D 0 
Note : (i) We cannot plot  on the paper. Infinity () is a symbol & not a number. It does not

E
obey the laws of elementry algebra. (ii)  +  =  (iii)  ×  = 
E a
R (iv) (a/) = 0 if a is finite (v) is not defined , if a  0.
0
F (vi) a b = 0 , if & only if a = 0 or b = 0 and a & b are finite.
6. The following strategies should be born in mind for evaluating the limits:
(a) Factorisation (b) Rationalisation or double rationalisation
(c) Use of trigonometric transformation ;
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