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BITSAT 2022 Paper
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Section:
Mathematics
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© examsnet.com
Question : 117
Total: 130
The value of
lim
x
→
0
(
1
+
x
)
1
x
−
e
+
1
2
e
x
x
2
is
11
24
e
−
11
24
e
e
24
None of these
Validate
Solution:
👈: Video Solution
Let
y
=
(
1
+
x
)
1
x
⇒
l
o
g
y
=
1
x
log
(
1
+
x
)
=
1
x
[
x
=
x
2
2
+
x
3
3
−
x
4
4
+
.
.
.
.
]
=
1
−
x
2
+
x
2
3
−
x
3
4
+
.
.
.
.
.
y
=
e
(
1
−
x
2
+
x
2
3
+
.
.
)
=
e
e
(
−
x
2
+
x
2
3.
.
.
)
=
e
[
1
+
(
−
x
2
+
x
2
3
−
.
.
.
.
)
+
1
2
!
(
−
x
2
+
x
2
3
−
.
.
.
)
2
+
.
.
.
]
y
−
e
+
1
2
e
x
=
e
x
2
[
1
3
+
0
(
x
)
+
1
2
(
−
1
2
+
0
(
x
)
)
2
+
.
.
.
]
[
∵
0
(
x
)
in terms containing
x
]
lim
y
−
e
+
1
2
e
x
x
2
=
e
[
1
3
+
1
8
]
=
11
24
e
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