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JEE Main Math Class 12 Limits and Derivatives Part 2 Questions
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© examsnet.com
Question : 32
Total: 56
If
lim
x
⟶
0
a
x
2
e
x
−
b
log
e
(
1
+
x
)
+
c
x
e
−
x
x
2
s
i
n
x
=
1
, then
16
(
a
2
+
b
2
+
c
2
)
is equal to_____
[31-Jan-2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
lim
x
⟶
0
ax
2
(
1
+
x
+
x
2
2
!
+
x
3
3
!
+
.
.
.
.
.
)
−
b
(
x
−
x
2
2
+
x
3
3
−
.
.
.
.
.
.
.
.
)
+
c
x
(
1
−
x
+
x
2
x
!
−
x
3
3
!
+
.
.
.
.
.
.
)
x
3
⋅
s
i
n
x
x
=
lim
x
⟶
∞
(
c
−
b
)
x
+
(
b
2
−
c
+
a
)
x
2
+
(
a
−
b
3
+
c
2
)
x
3
+
.
.
.
.
.
.
.
x
3
=
1
c
−
b
=
0
,
b
2
−
c
+
a
=
0
a
−
b
3
+
c
2
=
1
a
=
3
4
b
=
c
=
3
2
a
2
+
b
2
+
c
2
=
9
16
+
9
4
+
9
4
16
(
a
2
+
b
2
+
c
2
)
=
81
© examsnet.com
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