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Limits of Exponential Functions | Mathematics (Maths) Class 11 - Commerce PDF Download
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Page 1
LIMIT OF EXPONENTIAL FUNCTIONS
x 0
Lim
?
x
1 a
x
?
= ln a (a > 0). In particular
x 0
Lim
?
x
1 e
x
?
= 1
Ex.25 Evaluate
0 h
Lim
?
h
x ) h x (
x h x
? ?
?
(x > 0).
Sol.
0 h
Lim
?
h
e e
x n x ) h x ( n h x ? ?
?
? ?
=
0 h
Lim
?
x n x
e
?
?
?
?
?
?
?
?
?
?
? ? ?
h
1 e
x n x ) h x ( n h x ? ?
=
x
x
s
s 0 h 0
x h n (x h) x nx e 1
Lim . Lim
s h ? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
=
x
x
0 h
Lim
?
h
nx x
x
h
1 n nx h x ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
=
x
x
x/h
h 0 h 0
( x h x) (x h) h
Lim nx Lim n 1
h x x ? ?
? ?
? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
=
x
x
?
?
?
?
?
?
?
?
?
x
1
x 2
x n ?
Ex.26 Evaluate
0 x
Lim
?
x sin x
x 2 e e
x x
?
? ?
?
.
Sol. Let L =
0 x
Lim
?
3
3
x x
x
x sin x
. x
x 2 e e
?
? ?
?
Now ?
1
=
0 x
Lim
?
3
x
x sin x ?
=
1
6
(
use
x = 3 t
)
?
2
=
0 x
Lim
?
3
x x
x
x 2 e e ? ?
?
put x = 3
y
=
0 y
Limit
?
3
y 3 y 3
y 27
y 6 e e ? ?
?
=
L i m i t
y ? 0
3
y y 3 y y
y 27
y 6 e e 3 e e ) ( ) ( ? ? ? ?
? ?
?
?
?
?
?
?
?
?
?
2
8
3 y 2
0 y
9
1
) y 2 (
1 e
27
8
Limit
) (
? ?
2
=
9 27
8
2
?
? ? ?
2
=
3
1
Hence L =
2
1
?
?
= 2
Page 2
LIMIT OF EXPONENTIAL FUNCTIONS
x 0
Lim
?
x
1 a
x
?
= ln a (a > 0). In particular
x 0
Lim
?
x
1 e
x
?
= 1
Ex.25 Evaluate
0 h
Lim
?
h
x ) h x (
x h x
? ?
?
(x > 0).
Sol.
0 h
Lim
?
h
e e
x n x ) h x ( n h x ? ?
?
? ?
=
0 h
Lim
?
x n x
e
?
?
?
?
?
?
?
?
?
?
? ? ?
h
1 e
x n x ) h x ( n h x ? ?
=
x
x
s
s 0 h 0
x h n (x h) x nx e 1
Lim . Lim
s h ? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
=
x
x
0 h
Lim
?
h
nx x
x
h
1 n nx h x ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
=
x
x
x/h
h 0 h 0
( x h x) (x h) h
Lim nx Lim n 1
h x x ? ?
? ?
? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
=
x
x
?
?
?
?
?
?
?
?
?
x
1
x 2
x n ?
Ex.26 Evaluate
0 x
Lim
?
x sin x
x 2 e e
x x
?
? ?
?
.
Sol. Let L =
0 x
Lim
?
3
3
x x
x
x sin x
. x
x 2 e e
?
? ?
?
Now ?
1
=
0 x
Lim
?
3
x
x sin x ?
=
1
6
(
use
x = 3 t
)
?
2
=
0 x
Lim
?
3
x x
x
x 2 e e ? ?
?
put x = 3
y
=
0 y
Limit
?
3
y 3 y 3
y 27
y 6 e e ? ?
?
=
L i m i t
y ? 0
3
y y 3 y y
y 27
y 6 e e 3 e e ) ( ) ( ? ? ? ?
? ?
?
?
?
?
?
?
?
?
?
2
8
3 y 2
0 y
9
1
) y 2 (
1 e
27
8
Limit
) (
? ?
2
=
9 27
8
2
?
? ? ?
2
=
3
1
Hence L =
2
1
?
?
= 2
(4) (a)
0 x
Lim
?
(1 + x)
1/x
= e =
? ? x
Lim
x
x
1
1 ?
?
?
?
?
?
?
Note :
0 x
Lim
?
x
1
?n (1 + x) = 1
(b) Generalized formula for 1
?
form :
If
a x
Lim
?
f(x) = 1 and
a x
Lim
?
?? ?(x) = ? ?, then
a x
Lim
?
x a
Lim
(x) (x)[f(x) 1]
f(x) e
?
? ? ?
? ? ?
? ?
Ex.27 Evaluate
2
0 t
t
) t cos(sin n
lim
?
?
Sol.
2
2
2
2
0 t
2
0 t
t
2
t sin
sin 2
.
2
t sin
sin 2
2
) t (sin
sin 2 1 n
lim
t
) t cos(sin n
lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
=
2
1
t . 4
t sin
.
2
t sin
2
t sin
sin
. 2 lim
2
2
2
0 t
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Ex.28 Find
x cot 1
x tan ln
lim
4 / x
?
? ?
Sol. Put x = t + ?/4 ?
) 4 / t cot( 1
) 4 / t tan( ln
lim
0 t
? ? ?
? ?
?
=
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
4 / cot t cot
1 4 / cot t cot
1
t tan 1
t tan 1
ln
lim
0 t
=
t tan t
t tan 2
) t tan 1 ln(
lim
0 t
?
?
?
+
t tan 2
) t tan 1 )( t tan 1 ln(
lim
0 t
?
? ?
?
=
2
1
[1.1 + 1.1] =
2
2
= 1.
Ex.29 Evaluate
0 x
Lim
?
x . 1 e
x x 1 n
) (
) (
x
4 2
?
? ? ?
.
Sol.
0 x
Lim
?
2
4 2
x
x x 1 n ) ( ? ? ?
=
0 x
Lim
?
?n
2
x
1
4 2
) x x 1 ( ? ?
= ?n
0 x
Lim
?
) ( 1 x x 1
4 2
2
x
1
e
? ? ?
=
0 x
Lim
?
2
4 2
x
x x ?
=
0 x
Lim
?
(1 + x
2
) = 1
Ex.30 Evaluate :
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
bx 2
sec
2
2
ax 2
sin
Sol. ? =
0 x
Lim
?
x b 2
2
sec
2
x a 2
sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
x a 2
cos
x b 2
sec Limit
2 2
0 x
e
?
?
?
?
?
?
consider
0 x
Lim
?
bx 2
x a 2
cos
cos
?
?
?
?
=
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
? ?
x b 2 2
x a 2 2
sin
sin
=
0 x
Lim
?
x b 2 2
x a 2 2
?
?
?
?
?
?
?
?
=
0 x
Lim
?
x a 2
x a
?
?
x b
x b 2
?
?
=
b
a
? ? =
2
b
2
a
e
?
Page 3
LIMIT OF EXPONENTIAL FUNCTIONS
x 0
Lim
?
x
1 a
x
?
= ln a (a > 0). In particular
x 0
Lim
?
x
1 e
x
?
= 1
Ex.25 Evaluate
0 h
Lim
?
h
x ) h x (
x h x
? ?
?
(x > 0).
Sol.
0 h
Lim
?
h
e e
x n x ) h x ( n h x ? ?
?
? ?
=
0 h
Lim
?
x n x
e
?
?
?
?
?
?
?
?
?
?
? ? ?
h
1 e
x n x ) h x ( n h x ? ?
=
x
x
s
s 0 h 0
x h n (x h) x nx e 1
Lim . Lim
s h ? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
=
x
x
0 h
Lim
?
h
nx x
x
h
1 n nx h x ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
=
x
x
x/h
h 0 h 0
( x h x) (x h) h
Lim nx Lim n 1
h x x ? ?
? ?
? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
=
x
x
?
?
?
?
?
?
?
?
?
x
1
x 2
x n ?
Ex.26 Evaluate
0 x
Lim
?
x sin x
x 2 e e
x x
?
? ?
?
.
Sol. Let L =
0 x
Lim
?
3
3
x x
x
x sin x
. x
x 2 e e
?
? ?
?
Now ?
1
=
0 x
Lim
?
3
x
x sin x ?
=
1
6
(
use
x = 3 t
)
?
2
=
0 x
Lim
?
3
x x
x
x 2 e e ? ?
?
put x = 3
y
=
0 y
Limit
?
3
y 3 y 3
y 27
y 6 e e ? ?
?
=
L i m i t
y ? 0
3
y y 3 y y
y 27
y 6 e e 3 e e ) ( ) ( ? ? ? ?
? ?
?
?
?
?
?
?
?
?
?
2
8
3 y 2
0 y
9
1
) y 2 (
1 e
27
8
Limit
) (
? ?
2
=
9 27
8
2
?
? ? ?
2
=
3
1
Hence L =
2
1
?
?
= 2
(4) (a)
0 x
Lim
?
(1 + x)
1/x
= e =
? ? x
Lim
x
x
1
1 ?
?
?
?
?
?
?
Note :
0 x
Lim
?
x
1
?n (1 + x) = 1
(b) Generalized formula for 1
?
form :
If
a x
Lim
?
f(x) = 1 and
a x
Lim
?
?? ?(x) = ? ?, then
a x
Lim
?
x a
Lim
(x) (x)[f(x) 1]
f(x) e
?
? ? ?
? ? ?
? ?
Ex.27 Evaluate
2
0 t
t
) t cos(sin n
lim
?
?
Sol.
2
2
2
2
0 t
2
0 t
t
2
t sin
sin 2
.
2
t sin
sin 2
2
) t (sin
sin 2 1 n
lim
t
) t cos(sin n
lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
=
2
1
t . 4
t sin
.
2
t sin
2
t sin
sin
. 2 lim
2
2
2
0 t
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Ex.28 Find
x cot 1
x tan ln
lim
4 / x
?
? ?
Sol. Put x = t + ?/4 ?
) 4 / t cot( 1
) 4 / t tan( ln
lim
0 t
? ? ?
? ?
?
=
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
4 / cot t cot
1 4 / cot t cot
1
t tan 1
t tan 1
ln
lim
0 t
=
t tan t
t tan 2
) t tan 1 ln(
lim
0 t
?
?
?
+
t tan 2
) t tan 1 )( t tan 1 ln(
lim
0 t
?
? ?
?
=
2
1
[1.1 + 1.1] =
2
2
= 1.
Ex.29 Evaluate
0 x
Lim
?
x . 1 e
x x 1 n
) (
) (
x
4 2
?
? ? ?
.
Sol.
0 x
Lim
?
2
4 2
x
x x 1 n ) ( ? ? ?
=
0 x
Lim
?
?n
2
x
1
4 2
) x x 1 ( ? ?
= ?n
0 x
Lim
?
) ( 1 x x 1
4 2
2
x
1
e
? ? ?
=
0 x
Lim
?
2
4 2
x
x x ?
=
0 x
Lim
?
(1 + x
2
) = 1
Ex.30 Evaluate :
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
bx 2
sec
2
2
ax 2
sin
Sol. ? =
0 x
Lim
?
x b 2
2
sec
2
x a 2
sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
x a 2
cos
x b 2
sec Limit
2 2
0 x
e
?
?
?
?
?
?
consider
0 x
Lim
?
bx 2
x a 2
cos
cos
?
?
?
?
=
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
? ?
x b 2 2
x a 2 2
sin
sin
=
0 x
Lim
?
x b 2 2
x a 2 2
?
?
?
?
?
?
?
?
=
0 x
Lim
?
x a 2
x a
?
?
x b
x b 2
?
?
=
b
a
? ? =
2
b
2
a
e
?
Ex.31 Evaluate
? ? x
lim
( ? – 2 tan
–1
x) ln x
Sol. ?
?
?
?
?
? ? ?
?
? ?
0
0
nx
1
) x tan 2 (
lim
1
x
?
Applying L’ Hospital’s Rule=
x
1
) x (ln
1
x 1
2
lim
2
2
x
?
?
?
? ?
=
2
x
x 1
2
lim
?
? ?
(ln x)
2
=
x 2
x
1
. x ln 2 . x 2 ) x (ln 2
lim
2
x
?
? ?
=
x
x ln 2 ) x (ln
lim
2
x
?
? ?
=
x
x
2
x
1
. x ln 2
lim
x
?
? ?
=
2
x
x
1
lim
? ?
[2ln x + 3] = 0
x
1
? 0
Ex.32 Find a polynomial of least degree, such that
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
= e
2
.
Sol. Now,
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
exists only when
2
2
0 x
x
) x ( f x
lim
?
?
= 0 (i.e. it converts to 1
?
form).
So, the least degree in f(x) is of degree 2. i.e. f(x) = a
2
x
2
+ a
3
x
3
+ ......
Now,
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
= e
2
?
x
1
x
) x ( f x
lim
2
2
0 x
e
?
?
?
?
?
?
?
?
?
?
= e
2
?
3
2
0 x
x
) x ( f x
lim
e
?
?
= e
2
?
3
2
x
x
) x ( f x
lim
?
= 2 ?
3
3
3
2
2
2
x
x
..... x a x a x
lim
? ? ?
= 2
? a
2
= –1, a
3
= 2 and a
4
, a
5
....... are any arbitrary constants. Since, we want polynomial of least
degree.
Hence,f(x) = x
2
+ 2x
3
.
Ex.33 Evaluate
? ? x
Lim
x ? x
2
?n ?
?
?
?
?
?
?
x
1
1
.
Sol. Put
x =
y
1
? as x ? ?? ? ; y
?
0.
Hence ?
=
0 y
Lim
?
2
y
1
y
1
?
?n (1 + y) =
0 y
Lim
?
2
y
) y 1 ( n y ? ? ?
Put 1 + y = e
z
as y
?
0 , z
?
0
=
0 z
Lim
?
2 z
z
) (
) (
1 e
z 1 e
?
? ?
=
0 z
Lim
?
2
z
z
1 z e ? ?
Put z = 2
t
=
0 t
Lim
?
2
t 2
t 4
1 t 2 e ? ?
=
0 t
Lim
?
2
t 2 t
t 4
2 t 2 e 2 1 e ) ( ? ? ? ?
? =
4
1
+
2
l
? ? =
2
1
Page 4
LIMIT OF EXPONENTIAL FUNCTIONS
x 0
Lim
?
x
1 a
x
?
= ln a (a > 0). In particular
x 0
Lim
?
x
1 e
x
?
= 1
Ex.25 Evaluate
0 h
Lim
?
h
x ) h x (
x h x
? ?
?
(x > 0).
Sol.
0 h
Lim
?
h
e e
x n x ) h x ( n h x ? ?
?
? ?
=
0 h
Lim
?
x n x
e
?
?
?
?
?
?
?
?
?
?
? ? ?
h
1 e
x n x ) h x ( n h x ? ?
=
x
x
s
s 0 h 0
x h n (x h) x nx e 1
Lim . Lim
s h ? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
=
x
x
0 h
Lim
?
h
nx x
x
h
1 n nx h x ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
=
x
x
x/h
h 0 h 0
( x h x) (x h) h
Lim nx Lim n 1
h x x ? ?
? ?
? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
=
x
x
?
?
?
?
?
?
?
?
?
x
1
x 2
x n ?
Ex.26 Evaluate
0 x
Lim
?
x sin x
x 2 e e
x x
?
? ?
?
.
Sol. Let L =
0 x
Lim
?
3
3
x x
x
x sin x
. x
x 2 e e
?
? ?
?
Now ?
1
=
0 x
Lim
?
3
x
x sin x ?
=
1
6
(
use
x = 3 t
)
?
2
=
0 x
Lim
?
3
x x
x
x 2 e e ? ?
?
put x = 3
y
=
0 y
Limit
?
3
y 3 y 3
y 27
y 6 e e ? ?
?
=
L i m i t
y ? 0
3
y y 3 y y
y 27
y 6 e e 3 e e ) ( ) ( ? ? ? ?
? ?
?
?
?
?
?
?
?
?
?
2
8
3 y 2
0 y
9
1
) y 2 (
1 e
27
8
Limit
) (
? ?
2
=
9 27
8
2
?
? ? ?
2
=
3
1
Hence L =
2
1
?
?
= 2
(4) (a)
0 x
Lim
?
(1 + x)
1/x
= e =
? ? x
Lim
x
x
1
1 ?
?
?
?
?
?
?
Note :
0 x
Lim
?
x
1
?n (1 + x) = 1
(b) Generalized formula for 1
?
form :
If
a x
Lim
?
f(x) = 1 and
a x
Lim
?
?? ?(x) = ? ?, then
a x
Lim
?
x a
Lim
(x) (x)[f(x) 1]
f(x) e
?
? ? ?
? ? ?
? ?
Ex.27 Evaluate
2
0 t
t
) t cos(sin n
lim
?
?
Sol.
2
2
2
2
0 t
2
0 t
t
2
t sin
sin 2
.
2
t sin
sin 2
2
) t (sin
sin 2 1 n
lim
t
) t cos(sin n
lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
=
2
1
t . 4
t sin
.
2
t sin
2
t sin
sin
. 2 lim
2
2
2
0 t
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Ex.28 Find
x cot 1
x tan ln
lim
4 / x
?
? ?
Sol. Put x = t + ?/4 ?
) 4 / t cot( 1
) 4 / t tan( ln
lim
0 t
? ? ?
? ?
?
=
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
4 / cot t cot
1 4 / cot t cot
1
t tan 1
t tan 1
ln
lim
0 t
=
t tan t
t tan 2
) t tan 1 ln(
lim
0 t
?
?
?
+
t tan 2
) t tan 1 )( t tan 1 ln(
lim
0 t
?
? ?
?
=
2
1
[1.1 + 1.1] =
2
2
= 1.
Ex.29 Evaluate
0 x
Lim
?
x . 1 e
x x 1 n
) (
) (
x
4 2
?
? ? ?
.
Sol.
0 x
Lim
?
2
4 2
x
x x 1 n ) ( ? ? ?
=
0 x
Lim
?
?n
2
x
1
4 2
) x x 1 ( ? ?
= ?n
0 x
Lim
?
) ( 1 x x 1
4 2
2
x
1
e
? ? ?
=
0 x
Lim
?
2
4 2
x
x x ?
=
0 x
Lim
?
(1 + x
2
) = 1
Ex.30 Evaluate :
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
bx 2
sec
2
2
ax 2
sin
Sol. ? =
0 x
Lim
?
x b 2
2
sec
2
x a 2
sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
x a 2
cos
x b 2
sec Limit
2 2
0 x
e
?
?
?
?
?
?
consider
0 x
Lim
?
bx 2
x a 2
cos
cos
?
?
?
?
=
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
? ?
x b 2 2
x a 2 2
sin
sin
=
0 x
Lim
?
x b 2 2
x a 2 2
?
?
?
?
?
?
?
?
=
0 x
Lim
?
x a 2
x a
?
?
x b
x b 2
?
?
=
b
a
? ? =
2
b
2
a
e
?
Ex.31 Evaluate
? ? x
lim
( ? – 2 tan
–1
x) ln x
Sol. ?
?
?
?
?
? ? ?
?
? ?
0
0
nx
1
) x tan 2 (
lim
1
x
?
Applying L’ Hospital’s Rule=
x
1
) x (ln
1
x 1
2
lim
2
2
x
?
?
?
? ?
=
2
x
x 1
2
lim
?
? ?
(ln x)
2
=
x 2
x
1
. x ln 2 . x 2 ) x (ln 2
lim
2
x
?
? ?
=
x
x ln 2 ) x (ln
lim
2
x
?
? ?
=
x
x
2
x
1
. x ln 2
lim
x
?
? ?
=
2
x
x
1
lim
? ?
[2ln x + 3] = 0
x
1
? 0
Ex.32 Find a polynomial of least degree, such that
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
= e
2
.
Sol. Now,
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
exists only when
2
2
0 x
x
) x ( f x
lim
?
?
= 0 (i.e. it converts to 1
?
form).
So, the least degree in f(x) is of degree 2. i.e. f(x) = a
2
x
2
+ a
3
x
3
+ ......
Now,
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
= e
2
?
x
1
x
) x ( f x
lim
2
2
0 x
e
?
?
?
?
?
?
?
?
?
?
= e
2
?
3
2
0 x
x
) x ( f x
lim
e
?
?
= e
2
?
3
2
x
x
) x ( f x
lim
?
= 2 ?
3
3
3
2
2
2
x
x
..... x a x a x
lim
? ? ?
= 2
? a
2
= –1, a
3
= 2 and a
4
, a
5
....... are any arbitrary constants. Since, we want polynomial of least
degree.
Hence,f(x) = x
2
+ 2x
3
.
Ex.33 Evaluate
? ? x
Lim
x ? x
2
?n ?
?
?
?
?
?
?
x
1
1
.
Sol. Put
x =
y
1
? as x ? ?? ? ; y
?
0.
Hence ?
=
0 y
Lim
?
2
y
1
y
1
?
?n (1 + y) =
0 y
Lim
?
2
y
) y 1 ( n y ? ? ?
Put 1 + y = e
z
as y
?
0 , z
?
0
=
0 z
Lim
?
2 z
z
) (
) (
1 e
z 1 e
?
? ?
=
0 z
Lim
?
2
z
z
1 z e ? ?
Put z = 2
t
=
0 t
Lim
?
2
t 2
t 4
1 t 2 e ? ?
=
0 t
Lim
?
2
t 2 t
t 4
2 t 2 e 2 1 e ) ( ? ? ? ?
? =
4
1
+
2
l
? ? =
2
1
(5) BINOMIAL LIMIT :
a x
Lim
?
n n
n 1
x a
na
x a
?
?
?
?
Ex.34 Let
a x
Lim
?
a x
a x
x x
?
?
= ? , a > 0 &
a x
Lim
?
a x
x a
a x
?
?
= m , a > 0 . If ? = m then find the value of
'a'.
Sol. ?
=
a x
Lim
?
a x
e e
a n x x n x
?
?
? ?
=
? ?
) a n x n ( x
1 e e
) a n x n ( x a n x
? ?
? ? ?
?
?
?
.
a x
) a n x n ( x
?
? ? ?
= a
a
.
0 h
Lim
?
h
a
h
1 n ) h a ( ?
?
?
?
?
?
? ? ?
=
a
a
m =
a x
Lim
?
a x
e e
x n a a n x
?
?
? ?
=
a n x x n a a n x
x a
e e 1
Lim x n a a n x
?
?
? ?
?
? ?
?
? ? ?
? ?
.
a x
x n a a n x
?
? ? ?
= a
a
.
0 h
Lim
?
h
) h a ( n a a n ) h a ( ? ? ? ? ?
= a
a
0 h
Lim
?
? ?
?
?
?
?
?
?
? ?
h / a
a
h
1 n a n ? ?
= a
a
( ?n a ? 1)
Now ?
= m ? a = e
2
Ex.35 Evaluate
? ? n
Limit
?
?
?
?
?
?
? ? ? ?
?1 n n 2 3 2
2
x
sec .
2
x
tan ......
2
x
sec .
2
x
tan
2
x
sec .
2
x
tan x sec .
2
x
tan
where x
?
?
?
?
?
?
? ?
2
, 0 .
Sol. T
1
=
x cos
2
x
cos
2
x
sin
=
x cos
2
x
cos
2
x
x sin ?
?
?
?
?
?
?
= tan x
?
tan
2
x
Similarly T
2
= tan
2
x
?
tan
2
2
x
and T
3
= tan
2
2
x
?
tan
3
2
x
T
n
= tan
1 n
2
x
?
?
tan
n
2
x
S = tan x
?
tan
n
2
x
?
? ? n
Lim
S = tan x
Ex.36 Let f (x) =
?
?
?
? ?
n
1 n
n
3 1 n
n
3
x
sin 3 Lim
and g (x) = x – 4 f (x). Evaluate ? ?
x c o t
0 x
) x ( g 1 L i m ?
?
.
Sol. Using sin 3 ? = 3 sin ? – 4 sin
3
?
T
1
= sin
3
3
x
=
4
1
?
?
?
?
?
?
? x sin
3
x
sin 3
T
2
= 3sin
3
2
3
x
=
4
3
?
?
?
?
?
?
?
3
x
sin
3
x
sin 3
2
and so on........
? T
1
=
?
?
?
?
?
?
? x sin
3
x
sin 3
4
1
T
2
=
?
?
?
?
?
?
?
3
x
sin 3
3
x
sin 3
4
1
2
2
T
n
=
?
?
?
?
?
?
?
?
?
1 n
1 n
n
n
3
x
sin 3
3
x
sin 3
4
1
Page 5
LIMIT OF EXPONENTIAL FUNCTIONS
x 0
Lim
?
x
1 a
x
?
= ln a (a > 0). In particular
x 0
Lim
?
x
1 e
x
?
= 1
Ex.25 Evaluate
0 h
Lim
?
h
x ) h x (
x h x
? ?
?
(x > 0).
Sol.
0 h
Lim
?
h
e e
x n x ) h x ( n h x ? ?
?
? ?
=
0 h
Lim
?
x n x
e
?
?
?
?
?
?
?
?
?
?
? ? ?
h
1 e
x n x ) h x ( n h x ? ?
=
x
x
s
s 0 h 0
x h n (x h) x nx e 1
Lim . Lim
s h ? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
=
x
x
0 h
Lim
?
h
nx x
x
h
1 n nx h x ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
=
x
x
x/h
h 0 h 0
( x h x) (x h) h
Lim nx Lim n 1
h x x ? ?
? ?
? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ?
=
x
x
?
?
?
?
?
?
?
?
?
x
1
x 2
x n ?
Ex.26 Evaluate
0 x
Lim
?
x sin x
x 2 e e
x x
?
? ?
?
.
Sol. Let L =
0 x
Lim
?
3
3
x x
x
x sin x
. x
x 2 e e
?
? ?
?
Now ?
1
=
0 x
Lim
?
3
x
x sin x ?
=
1
6
(
use
x = 3 t
)
?
2
=
0 x
Lim
?
3
x x
x
x 2 e e ? ?
?
put x = 3
y
=
0 y
Limit
?
3
y 3 y 3
y 27
y 6 e e ? ?
?
=
L i m i t
y ? 0
3
y y 3 y y
y 27
y 6 e e 3 e e ) ( ) ( ? ? ? ?
? ?
?
?
?
?
?
?
?
?
?
2
8
3 y 2
0 y
9
1
) y 2 (
1 e
27
8
Limit
) (
? ?
2
=
9 27
8
2
?
? ? ?
2
=
3
1
Hence L =
2
1
?
?
= 2
(4) (a)
0 x
Lim
?
(1 + x)
1/x
= e =
? ? x
Lim
x
x
1
1 ?
?
?
?
?
?
?
Note :
0 x
Lim
?
x
1
?n (1 + x) = 1
(b) Generalized formula for 1
?
form :
If
a x
Lim
?
f(x) = 1 and
a x
Lim
?
?? ?(x) = ? ?, then
a x
Lim
?
x a
Lim
(x) (x)[f(x) 1]
f(x) e
?
? ? ?
? ? ?
? ?
Ex.27 Evaluate
2
0 t
t
) t cos(sin n
lim
?
?
Sol.
2
2
2
2
0 t
2
0 t
t
2
t sin
sin 2
.
2
t sin
sin 2
2
) t (sin
sin 2 1 n
lim
t
) t cos(sin n
lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
=
2
1
t . 4
t sin
.
2
t sin
2
t sin
sin
. 2 lim
2
2
2
0 t
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Ex.28 Find
x cot 1
x tan ln
lim
4 / x
?
? ?
Sol. Put x = t + ?/4 ?
) 4 / t cot( 1
) 4 / t tan( ln
lim
0 t
? ? ?
? ?
?
=
?
?
?
?
?
?
? ?
? ?
?
?
?
?
?
?
?
?
?
?
4 / cot t cot
1 4 / cot t cot
1
t tan 1
t tan 1
ln
lim
0 t
=
t tan t
t tan 2
) t tan 1 ln(
lim
0 t
?
?
?
+
t tan 2
) t tan 1 )( t tan 1 ln(
lim
0 t
?
? ?
?
=
2
1
[1.1 + 1.1] =
2
2
= 1.
Ex.29 Evaluate
0 x
Lim
?
x . 1 e
x x 1 n
) (
) (
x
4 2
?
? ? ?
.
Sol.
0 x
Lim
?
2
4 2
x
x x 1 n ) ( ? ? ?
=
0 x
Lim
?
?n
2
x
1
4 2
) x x 1 ( ? ?
= ?n
0 x
Lim
?
) ( 1 x x 1
4 2
2
x
1
e
? ? ?
=
0 x
Lim
?
2
4 2
x
x x ?
=
0 x
Lim
?
(1 + x
2
) = 1
Ex.30 Evaluate :
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
bx 2
sec
2
2
ax 2
sin
Sol. ? =
0 x
Lim
?
x b 2
2
sec
2
x a 2
sin
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
x a 2
cos
x b 2
sec Limit
2 2
0 x
e
?
?
?
?
?
?
consider
0 x
Lim
?
bx 2
x a 2
cos
cos
?
?
?
?
=
0 x
Lim
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
? ?
x b 2 2
x a 2 2
sin
sin
=
0 x
Lim
?
x b 2 2
x a 2 2
?
?
?
?
?
?
?
?
=
0 x
Lim
?
x a 2
x a
?
?
x b
x b 2
?
?
=
b
a
? ? =
2
b
2
a
e
?
Ex.31 Evaluate
? ? x
lim
( ? – 2 tan
–1
x) ln x
Sol. ?
?
?
?
?
? ? ?
?
? ?
0
0
nx
1
) x tan 2 (
lim
1
x
?
Applying L’ Hospital’s Rule=
x
1
) x (ln
1
x 1
2
lim
2
2
x
?
?
?
? ?
=
2
x
x 1
2
lim
?
? ?
(ln x)
2
=
x 2
x
1
. x ln 2 . x 2 ) x (ln 2
lim
2
x
?
? ?
=
x
x ln 2 ) x (ln
lim
2
x
?
? ?
=
x
x
2
x
1
. x ln 2
lim
x
?
? ?
=
2
x
x
1
lim
? ?
[2ln x + 3] = 0
x
1
? 0
Ex.32 Find a polynomial of least degree, such that
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
= e
2
.
Sol. Now,
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
exists only when
2
2
0 x
x
) x ( f x
lim
?
?
= 0 (i.e. it converts to 1
?
form).
So, the least degree in f(x) is of degree 2. i.e. f(x) = a
2
x
2
+ a
3
x
3
+ ......
Now,
x / 1
2
2
0 x
x
) x ( f x
1 lim
?
?
?
?
?
?
?
?
?
?
?
= e
2
?
x
1
x
) x ( f x
lim
2
2
0 x
e
?
?
?
?
?
?
?
?
?
?
= e
2
?
3
2
0 x
x
) x ( f x
lim
e
?
?
= e
2
?
3
2
x
x
) x ( f x
lim
?
= 2 ?
3
3
3
2
2
2
x
x
..... x a x a x
lim
? ? ?
= 2
? a
2
= –1, a
3
= 2 and a
4
, a
5
....... are any arbitrary constants. Since, we want polynomial of least
degree.
Hence,f(x) = x
2
+ 2x
3
.
Ex.33 Evaluate
? ? x
Lim
x ? x
2
?n ?
?
?
?
?
?
?
x
1
1
.
Sol. Put
x =
y
1
? as x ? ?? ? ; y
?
0.
Hence ?
=
0 y
Lim
?
2
y
1
y
1
?
?n (1 + y) =
0 y
Lim
?
2
y
) y 1 ( n y ? ? ?
Put 1 + y = e
z
as y
?
0 , z
?
0
=
0 z
Lim
?
2 z
z
) (
) (
1 e
z 1 e
?
? ?
=
0 z
Lim
?
2
z
z
1 z e ? ?
Put z = 2
t
=
0 t
Lim
?
2
t 2
t 4
1 t 2 e ? ?
=
0 t
Lim
?
2
t 2 t
t 4
2 t 2 e 2 1 e ) ( ? ? ? ?
? =
4
1
+
2
l
? ? =
2
1
(5) BINOMIAL LIMIT :
a x
Lim
?
n n
n 1
x a
na
x a
?
?
?
?
Ex.34 Let
a x
Lim
?
a x
a x
x x
?
?
= ? , a > 0 &
a x
Lim
?
a x
x a
a x
?
?
= m , a > 0 . If ? = m then find the value of
'a'.
Sol. ?
=
a x
Lim
?
a x
e e
a n x x n x
?
?
? ?
=
? ?
) a n x n ( x
1 e e
) a n x n ( x a n x
? ?
? ? ?
?
?
?
.
a x
) a n x n ( x
?
? ? ?
= a
a
.
0 h
Lim
?
h
a
h
1 n ) h a ( ?
?
?
?
?
?
? ? ?
=
a
a
m =
a x
Lim
?
a x
e e
x n a a n x
?
?
? ?
=
a n x x n a a n x
x a
e e 1
Lim x n a a n x
?
?
? ?
?
? ?
?
? ? ?
? ?
.
a x
x n a a n x
?
? ? ?
= a
a
.
0 h
Lim
?
h
) h a ( n a a n ) h a ( ? ? ? ? ?
= a
a
0 h
Lim
?
? ?
?
?
?
?
?
?
? ?
h / a
a
h
1 n a n ? ?
= a
a
( ?n a ? 1)
Now ?
= m ? a = e
2
Ex.35 Evaluate
? ? n
Limit
?
?
?
?
?
?
? ? ? ?
?1 n n 2 3 2
2
x
sec .
2
x
tan ......
2
x
sec .
2
x
tan
2
x
sec .
2
x
tan x sec .
2
x
tan
where x
?
?
?
?
?
?
? ?
2
, 0 .
Sol. T
1
=
x cos
2
x
cos
2
x
sin
=
x cos
2
x
cos
2
x
x sin ?
?
?
?
?
?
?
= tan x
?
tan
2
x
Similarly T
2
= tan
2
x
?
tan
2
2
x
and T
3
= tan
2
2
x
?
tan
3
2
x
T
n
= tan
1 n
2
x
?
?
tan
n
2
x
S = tan x
?
tan
n
2
x
?
? ? n
Lim
S = tan x
Ex.36 Let f (x) =
?
?
?
? ?
n
1 n
n
3 1 n
n
3
x
sin 3 Lim
and g (x) = x – 4 f (x). Evaluate ? ?
x c o t
0 x
) x ( g 1 L i m ?
?
.
Sol. Using sin 3 ? = 3 sin ? – 4 sin
3
?
T
1
= sin
3
3
x
=
4
1
?
?
?
?
?
?
? x sin
3
x
sin 3
T
2
= 3sin
3
2
3
x
=
4
3
?
?
?
?
?
?
?
3
x
sin
3
x
sin 3
2
and so on........
? T
1
=
?
?
?
?
?
?
? x sin
3
x
sin 3
4
1
T
2
=
?
?
?
?
?
?
?
3
x
sin 3
3
x
sin 3
4
1
2
2
T
n
=
?
?
?
?
?
?
?
?
?
1 n
1 n
n
n
3
x
sin 3
3
x
sin 3
4
1
Limit – Nirmaan TYCRP
? f (x) =
?
?
?
?
?
?
?
? ?
x sin
3
x
sin 3
4
1
Lim
n
n
n
=
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
x sin
3
x
3
x
sin x
Lim
4
1
n
n
n =
4
1
(x – sin x)
g (x) = x – 4 f (x) = sin x
now
x cot
0 x
) x sin 1 ( Lim ?
?
=
) x )(sin x (cot Lim
0 x
e
?
= e
Ex.37 Evaluate
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
? ?
2
r
1
r r
cot lim
3
n
1 r
1
n
Sol. Here,
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
? ?
2
r
1
r r
cot lim
3
n
1 r
1
n
=
?
?
?
?
?
?
? ?
?
?
?
? ?
4 2
n
1 r
1
n
r r 1
r 2
tan lim
=
?
?
?
?
?
?
?
?
? ? ?
?
?
?
? ?
) r r )( r r ( 1
r 2
tan lim
2 2
n
1 r
1
n
=
?
?
?
?
?
?
?
?
? ? ?
? ?
?
?
?
? ?
) r r )( r r ( 1
) r r )( r r (
tan lim
2 2
2 2 n
1 r
1
n
= ?
?
? ?
? ?
? ? ?
n
1 r
2 1 2 1
n
)} r r ( tan ) r r ( {tan lim
=
? ? n
lim
[(tan
–1
2 – tan
–1
0) + (tan
–1
6 – tan
–1
2) + (tan
–1
12 – tan
–1
6)
=
? ? n
lim
{tan
–1
(n
2
+ n) – tan
–1
(0)} = tan
–1
( ?) –tan
–1
(0) =
2
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? n
1 r
3
1
n
2 2
r
1
r r
cot lim
Ex.38 Evaluate
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
? ?
? ?
n
n
2
2
1
1
n
a
1 a
........
a
1 a
a
1 a
lim
, where a
1
= 1 and a
n
= n (1 + a
n – 1
) ? n ? 2.
Sol.
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
? ?
? ?
n
n
2
2
1
1
n
a
1 a
......
a
1 a
a
1 a
lim
We know, a
n – 1
+ 1 =
n
a
n
...(i)
n 2 1
1 n 4 3 2
n
a ..... a . a
1
1 n
a
.....
4
a
3
a
2
a
lim ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
=
)! 1 n (
a
lim
1 n
n
?
?
? ?
=
! n
a 1
lim
n
n
?
? ?
{using (ii)}
=
?
?
?
?
?
?
?
? ?
! n
a
! n
1
lim
n
n
=
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
)! 1 n (
a
)! 1 n (
1
! n
1
lim
1 n
n
{using (i)}
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FAQs on Limits of Exponential Functions - Mathematics (Maths) Class 11 - Commerce
| 1. What are the limits of exponential functions? |  |
Ans. The limits of exponential functions depend on the value of the base. If the base is greater than 1, the exponential function will approach positive infinity as x approaches positive infinity. Conversely, if the base is between 0 and 1, the exponential function will approach 0 as x approaches positive infinity.
| 2. Is there a limit to how large an exponential function can grow? | |
Ans. In theory, exponential functions can grow infinitely large as the input value increases. However, in practical terms, there are limitations due to factors such as computational power, available resources, and physical constraints. Eventually, exponential growth may become unsustainable or restricted by these limitations.
| 3. Can exponential functions have negative limits? | |
Ans. No, exponential functions with positive bases will never have a negative limit. As x approaches negative infinity, the function will either approach 0 or positive infinity, depending on the base. Negative limits are not possible for exponential functions with positive bases.
| 4. Do all exponential functions have limits at infinity? | |
Ans. No, not all exponential functions have limits at infinity. For example, if the base is equal to 1, the exponential function will remain constant and not approach any specific value as x approaches infinity. Additionally, exponential functions with negative bases will oscillate between positive and negative values, without approaching any limit.
| 5. Can exponential functions have limits at a finite value? | |
Ans. Yes, exponential functions can have limits at finite values. For example, if the base is equal to 1, the exponential function will have a limit of 1 as x approaches any finite value. Similarly, exponential functions with a base between 0 and 1 will approach 0 as x approaches any finite value.
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Dec 22, 2024 Last updated |
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