MATHS
Limit of Function
Definition : Limit of a function f(x) is said to exist, as x a when, im f (a h) h0
=
(Left hand limit)
im
h0
f (a + h) = Finite
(Right hand limit)
Note that we are not interested in knowing about what happens at x = a. Also note that if L.H.L. & R.H.L. are both tending towards ' ' or ‘–’, then it is said to be infinite limit. Remember, ‘x a’ means that x is approaching to ‘a’ but not equal to ‘a’.
Fundamental theorems on limits: im f (x) = and im g (x) = m. If & m are finite, then: Let x a x a (A)
im { f (x) ± g (x) } = ± m x a
(B)
im { f(x). g(x) } = . m x a
(C)
im f ( x) = , provided m x a g ( x) m
(D)
im k f(x) = k im f(x) = k; where k is a constant. x a x a
(E)
im f (g(x)) = f im g ( x ) = f (m); provided f is continuous at g (x) = m. x a x a
0
Example # 1 : Evaluate the following limits : -
Solution :
(i)
im (x + 2) x 2
(i)
im (x + 2) = 2 + 2 = 4 x + 2 being a polynomial in x, its limit as x 2 is given by x 2
(ii)
im cos (sin x) = cos im sin x = cos 0 = 1 x 0 x 0
(ii)
im cos (sin x) x 0
Self practice problems Evaluate the following limits : -
(1)
im x(x – 1) x 2
Answers :
(1)
(2)
2
(2)
2 im x 4 x 2 x2
2
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MATHS Indeterminate forms : If on putting x = a in f(x), any one of
0 , , 0 , º, 0º,1form is obtained, then the limit 0
has an indeterminate form. All the above forms are interchangeable, i.e. we can change one form to other by suitable substitutions etc.
im f(x) may exist. In such cases x a Consider f(x) =
x2 4 im x2 – 4 = 0 and im x – 2 = 0 . Here x 2 x 2 x2
im f(x) has an indeterminate form of the type 0 . x 2 0
im n x has an indeterminate form of type . x
x
im (1 + x)1/x is an indeterminate form of the type 1 x 0 NOTE : (i) (ii)
+= x =
(iii)
a = 0, if a is finite.
(iv)
a is not defined for any a R. 0
(v)
2 im x is an indeterminate form whereas im [ x ] is not an indeterminate form (where [.] x 0 x x 0 x2
represents greatest integer function) Students may remember these forms alongwith the prefix ‘tending to’ i.e.
exactly zero tending to zero is an indeterminate form whereas tending to zero is not an indeterminate tending to zero
form, its value is zero. similarly (tending to one) tending to is indeterminate form whereas (exactly one) tending to is not an indeterminate form, its value is one.
To evaluate a limit, we must always put the value where ' x ' is approaching to in the function. If we get a determinate form, then that value becomes the limit otherwise if an indeterminate form comes, we have to remove the indeterminancy, once the indeterminancy is removed the limit can be evaluated by putting the value of x, where it is approaching.
Methods of removing indeterminancy Basic methods of removing indeterminancy are (A) Factorisation (B) Rationalisation (C) Using standard limits (D) Substitution (E) Expansion of functions.
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MATHS Factorisation method :We can cancel out the factors which are leading to indeterminancy and find the limit of the remaining expression. x 2 2x 3
Example # 2 : (i)
im
Solution :
(i)
im
(ii)
2 (2 x 3) 1 im 3 x 2 x 2 x 3 x 2 2x
x 3
(ii)
x 4x 3 2
x 2 2x 3
x 3
x 4x 3 2
im = x 3
2 (2 x 3) 1 im 3 x 2 x 2 x 3 x 2 2x
( x 3)( x 1) =2 ( x 3)( x 1)
1 2 (2 x 3 ) im = x 2 x 2 x( x 1)( x 2) x( x 1) 2( 2 x 3 ) im = x 2 x( x 1)( x 2)
x 2 5x 6 im = x 2 x( x 1)( x 2) ( x 2)( x 3) x3 1 im im = x = x =– 2 x( x 1)( x 2) 2 x( x 1) 2
Rationalisation method :We can rationalise the irrational expression in numerator or denominator or in both to remove the indeterminancy. Example # 3 : Evaluate :
Solution :
(i)
im
(i)
im
4 15x 1
x 1
2 3x 1
1 x 1 x x
(4 15x 1)(2 3x 1)(4 15x 1) (2 3x 1)(4 15x 1)(2 3x 1)
(15 15x) 2 3x 1 5 × = (3 3x) 4 15x 1 2
The form of the given limit is
x 0
x 0
2 3x 1
= xim 1
im
im
4 15x 1
x 1
= xim 1
(ii)
(ii)
1 x 1 x x
0 when x 0. Rationalising the numerator, we get 0
1 x 1 x 1 x 1 x im = x 0 x 1 x 1 x (1 x ) (1 x ) 2x im = im = x 0 x ( 1 x 1 x ) x 0 x ( 1 x 1 x )
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MATHS 2 im = x 0 1 x 1 x
=
2 =1 2
Self practice problems Evaluate the following limits : (3)
(5)
(7)
im (2x 3) x 1 x 1 2x 2 x 3
im
h0
im
x 0
Answers :
xh x h
im 1 (sin x )1/ 3 2
(4)
x
(6)
im
1 (sin x )2 / 3 x b ab
x a
x 2 a2
x 4 x x
(3)
1 10
1 2
(4)
1 (5)
2 x
1 (6)
4a a b
(7)
0
Standard limits : (a)
(i)
im sinx = im tanx = 1 x 0 x 0 x x
(ii)
1 1 im tan x = im sin x = 1 x 0 x 0 x x
(iii)
im (1 x ) x = e x 0
;
xim (1 ax ) x e a 0
(iv)
im 1 1 = e ; x x
im 1 a e a x x
(v)
x im e 1 = 1 x 0 x
(vi)
im n(1 x ) = 1 x 0 x
(vii)
n n im x a = nan – 1 x a xa
1
1
x
(b)
[ Where x is measured in radians ]
;
x
x im a 1 = log a = na x 0 e x
,a > 0
If f(x) 0, when x a, then (i)
im
sin f ( x ) =1 f (x)
(ii)
im cos f(x) = 1 x a
(iii)
im
tan f ( x ) =1 f ( x)
(iv)
im
ef(x) 1 =1 f (x)
(vi)
im
n(1 f ( x )) =1 f ( x)
(v)
x a
x a
f(x) 1 im b = n b, (b > 0) x a f (x)
x a
x a
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MATHS 1
(vii)
(c)
im (1 f ( x )) f ( x ) e x a
im f(x) = A > 0 and im (x) = B(a finite quantity), then im [f(x)](x) = AB. x a x a x a
n im (1 x ) 1 Example # 4 : Evaluate : x 0 x
Solution :
n n im (1 x ) 1 = im (1 x ) 1 = n x 0 x 0 (1 x ) 1 x
3x im e 1 Example # 5 : Evaluate : x 0 x/2
Solution :
3x 3x im e 1 = im 2 × 3 e 1 = 6. x 0 x 0 3x x/2
im tan x sin x Example # 6 : Evaluate : x 0 x3
Solution :
x tan x . 2 sin2 im tan x sin x = im tan x(1 cos x ) = im 2 x 0 x 0 x 0 x3 x3 x3
im 1 . tan x = x 0 2 x
x sin 2 . x 2
2
=
1 . 2
im sin 2x Example # 7 : Evaluate : x 0 sin 3 x Solution :
3x sin 2x 2 3x sin 2x 2x im . . = x = 2xim . . 3xim 0 2x 0 0 3 x sin 3 x 2x 3 sin 3 x
im sin 2x x 0 sin 3 x
=1.
sin 3 x 2 2 2 × im = ×1= 3 x 0 3x 3 3 3
im 1 2 Example # 8 : Evaluate : x x
x
x
Solution :
2 im 1 2 = xim x .x = e2. e x x
x 3 im e e (ii) x 3 x3 Put y = x – 3. So, as x 3 y 0. Thus
Example # 9 : Evaluate : Solution :
(i)
(i)
x 3 im e e x 3 x3
3 y e3 im e = y 0 y
x im x(e 1) x 0 1 cos x
3 y 3 y im e . e e = e3 im e 1 = e3 . 1 = e3 = y 0 y 0 y y
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MATHS x im x(e 1) x 0 1 cos x
(ii)
x im x(e 1) = 1 . im = x 0 2 x 0 x 2 sin 2 2
ex 1 x2 . x x sin2 2
= 2.
Self practice problems Evaluate the following limits : (8)
im sin 2x x 0 x
(10)
im
(12)
x
x
4
(9)
1 sin 2 x 4x b
im (1 + a2)x sin
Answers :
(1 a 2 ) x
(8)
2
(12)
b
im x 0
8 x8
x2 x2 x2 x2 1 – cos – cos cos cos 2 4 2 4
(11)
x x im 5 9 x 0 x
(10)
does not exist
, where a 0
(9)
1 32
(11)
n
5 9
Use of substitution in solving limit problems Sometimes in solving limit problem we convert xim f(x) into him f(a + h) or him 0 f(a – h) according as a 0 need of the problem. (here h is approaching to zero.)
1 tan x
Example # 10 : Evaluate
im
Solution :
+h 4
Put
x=
x
x
4
4
1 2 sin x
h0
1 tan h 4 im h0 1 2 sin h 4
2 tan h 1 tan h 1 tan h 1 tan h im im = h = h h h 0 1 sin h cos h 0 2 h 2 sin 2 sin cos 2 2 2 1
tanh 2 1 1 h 2 2 tan h im im = h = h = = 2. h 0 0 ( 1 tanh) ( 1 tanh) 1 h h h sin 2 sin sin cos 2 sin h cos h 2 2 2 h 2 2 2
Limits using expansion (a)
ax 1
x n a x 2 n 2 a x 3 n 3 a ........., a 0 1! 2! 3!
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MATHS x x2 x3 ...... 1! 2! 3!
(b)
e x 1
(c)
n (1+x) = x
(d)
sin x x
x3 x5 x7 ..... 3! 5! 7!
(e)
cosx 1
x2 x 4 x6 ..... 2! 4! 6!
(f)
tan x = x
(g)
tan-1x = x
x3 x5 x7 .... 3 5 7
(h)
sin-1x = x
12 3 12.3 2 5 12.3 2.5 2 7 x x x ..... 3! 5! 7!
(i)
sec -1x = 1
x 2 5 x 4 61x 6 ...... 2! 4! 6!
(j)
for |x| < 1, n R; (1 + x)n = 1 + nx +
(k)
1 x 11 2 (1 x ) x = e 1 2 24 x .......... ...
x2 x3 x4 ........., for 1 x 1 2 3 4
x 3 2x 5 ...... 3 15
n(n 1) n(n 1)(n 2) 2 x + x 3 + ............ 1. 2 1. 2 . 3
x im e 1 x Example # 11 : Evaluate x 0 x2
x2 1 x ...... 1 x e 1 x 1 2 ! im im = x = x0 0 2 2 2 x x x
Solution :
im tan x sin x Example # 12 : Evaluate x 0 x3
Solution :
x3 x3 x ....... x ...... tan x sin x 3 3 ! im im = 1 + 1 = 1. = x 3 x0 0 3 3 6 2 x x
1
7 x ) 3 2 Example # 13 : Evaluate xim 1 x 1 Solution :
Put x = 1 + h 1
3 im (8 h) 2 h0 h
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MATHS 2 1 1 h 1 1 h 3 3 8 2 1 . ....... 1 1 1. 2 h 3 3 8 2 . 1 2 8 im im im 2 × 1 = 1 = h = = h 0 h0 0 h 24 12 h
n(1 x ) sin x
im Example # 14 : Evaluate x 0
Solution :
x2 2
x tan x sin x
2 3 3 5 2 x x x ..... x x x ..... x x2 2 n(1 x ) sin x 2 3 3! 5! 1 1 1 im 2 = im = + = x0 x0 tan x sin x 3 6 2 x tan x sin x x3 . . x x
1
x im e (1 x ) Example # 15 : Evaluate x 0 tan x
Solution :
im e (1 x ) tan x
x0
1 x
x e e 1 ...... 2 = im e × x = e im = x 0 x0 tan x 2 2 tan x
Example # 16 : Find the values of a,b and c so that
Solution :
ae x b cos x ce x =2 x 0 x sin x
im
ae x b cos x ce x =2 x 0 x sin x
im
.....(1)
at x 0 numerator must be equal to zero a–b+c=0 b=a+c
.....(2)
From (1) & (2),
ae x (a c ) cos x ce x =2 x 0 x sin x
im
x x2 x3 x2 x4 a 1 ...... – (a c ) 1 ...... c 2! 4! im 1 ! 2 ! 3 ! x 0 3 5 x x x x ..... 3 ! 5 !
2 3 1 x x x ...... 1! 2 ! 3 ! =2
(a c ) x (a c ) (a c ) ..... x 3! im =2 x 0 2 4 1 x x ...... 3! 5!
Since R.H.S is finite,
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MATHS
a–c=0
a = c, then
0 2a 0 .... =2 1
a = 1 then c = 1 From (2), b=a+c=1+1=2
Limit when x In these types of problems we usually cancel out the greatest power of x common in numerator and denominator both. Also sometime when x , we use to substitute y =
1 and in this case y 0+. x
im x sin 1 Example # 17 : Evaluate x x
Solution :
im Example # 18 : Evaluate x
Solution :
sin
1 im im = x x x sin x
1 x
=1
x2 2x 3
im x 2 im x 2x 3 x
im Example # 19 : Evaluate x
1 x
2 1 x = . 3 2 2 x 1
x 2 4x 5 3x 2 x 3 2
1 4 5 2 3 x im x 4 x 5 = im x x x x 3 2 3x 2 x 3 2 1 3 x x 2
Solution :
im Example # 20 : Evaluate x
Solution :
im
x
= tim 0
3x 2 2 x2
3x 2 2 x2
3 2t 2 . ( 1 2 t ) t
=0
(Put x = –
1 , x – t
t 0 +)
1 t2
= tim 0
t 3 2t 2 (1 2t ) | t |
=
3 =– 1
3.
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MATHS Some important notes : (i)
im nx = 0 x x
(iv)
im
x
nx n x
(ii)
=0
(v)
im
x
x ex
=0
(iii)
n im x = 0 x e x
im x(nx)n = 0
x 0
As x , n x increases much slower than any (positive) power of x where as ex increases much faster than any (positive) power of x. (vi)
im (1 h)n = 0 and im (1 + h)n where h 0+. n
n
1000 im x Example # 21 : Evaluate x ex
Solution :
1000 im x =0 x ex
Limits of form 1 , 0 0 , 0
(A)
All these forms can be converted into (a)
0 form in the following ways 0
If x 1, y , then z = (x)y is of 1 form n z = y n x
n z =
nx 1 y
0 form 0
1 As y y 0 and x 1 nx 0 (b)
If x 0, y 0, then z = x y is of (00) form n z = y n x
(c)
y 1 nx
0 form 0
If x , y 0, then z = x y is of ()0 form n z = y n x
(B)
n z =
n z =
y 1 nx
0 form 0
(1) type of problems can be solved by the following method 1
(a)
im (1 x ) x = e x 0
(b)
im [f(x)]g(x) ; where f(x) 1 ; g(x) as x a x a
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MATHS im 1 f ( x ) 1 = x a
1 im [1 ( f ( x ) 1)] f ( x )1 = x a
1 { f ( x ) 1} . g( x ) f ( x ) 1
( f ( x )1) g( x )
im [ f ( x )1] g( x )
= e xa
2x 2 1 im Example # 22 : Evaluate x 2x 2 3 Solution :
4 x 2
Since it is in the form of 1
2 im 2x 1 x 2x 2 3
4 x 2
= exp im x
2x 2 1 2x 2 3 ( 4 x 2 2) = e–8 2 2x 3
Example # 23 : Evaluate im (tan x)tan 2x x
Solution :
4
Since it is in the form of 1 so
im x
4
2
= e
im 2 a Example # 24 : Evaluate x a x
im 2 a x a x
Solution :
tan
tan
x 2a
(tan x)tan 2x = e tan / 4 1(1 tan / 4 )
im (tan x 1) tan 2 x
im (tan x 1)
x
x
4
= e–1 =
= e
4
2 tan x 1 tan 2 x
1 e
.
x 2a
put
im 1 h h 0 ( a h) h im 2a h h0 tan 2a e
2a . ah
x=a+h
h tan 2 2a
im 1 h h 0 ah
= e
–
h cot 2a
im cot
= e
h0
h h . 1 1 2a a h
2
im x x Example # 25 : Evaluate x0 Solution :
im x x Let y = x0
n y
im x n x = im – = x0 x 0
1 x = 0, as 1 x
n
1 x
y=1
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MATHS Sandwitch theorem or squeeze play theorem: Suppose that f(x) g(x) h(x) for all x in some open interval containing a, except possibly at x = a itself. Suppose also that
im f(x) = = im h(x), x a x a im g(x) = . Then x a
h g f
im [ x ] [2x ] [3 x ] .... [nx ] , where [.] denotes greatest integer function. Example # 26 : Evaluate n n2 Solution : We know that, x – 1 < [x] x 2x – 1 < [2x] 2x 3x – 1 < [3x] 3x . . . . . . nx – 1 < [nx] nx (x + 2x + 3x + .... + nx) – n < [x] + [2x] + ..... +[nx] (x + 2x + .... + nx) xn(n 1) –n< 2
[r x] r 1
x.n(n 1) 2
x 1 1 1 im [ x ] [2x ] .... [nx ] im x 1 1 – < n n 2 n n 2 n n2
im
n
n
x im [ x ] [2x ] .... [nx ] x < n 2 2 n2
[ x ] [2x ] .... [nx ]
im
n
n
2
=
x 2
Aliter We know that [x] = x – {x} n
[r x] r 1
= [x] + [2x] + .... + [nx] = (x + 2x + 3x + ... + nx) – ({x} + {2x} + .... + {nx})
xn(n 1) – ({x} + {2x} + .. + {nx}) 2
=
1 n
2
n
[r x] = r 1
{ x } {2x } .... {nx } x 1 1 – 2 n n2 n
Since, 0 {rx} < 1,
0
{r x}
r 1
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MATHS n
im
n
{rx} r 1
n
=0
2 n
n
im
n
[rx]
r 1
n
2
im x 1 1 – im = n n 2 n
{rx} r 1
n
2
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n
im
n
[rx]
r 1
n2
=
x 2
13
