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Standard XII
Mathematics
Algebra of Limits
Question
Evaluate:
lim
h
→
0
(
1
h
3
√
8
+
h
−
1
2
h
)
1
12
−
4
3
−
1
48
−
16
3
A
−
16
3
B
−
1
48
C
1
12
D
−
4
3
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Solution
Verified by Toppr
Let
L
=
lim
h
→
0
(
1
h
3
√
8
+
h
−
1
2
h
)
=
lim
h
→
0
1
h
(
1
3
√
8
+
h
−
1
2
)
Let
a
=
3
√
8
+
h
⇒
a
3
=
8
+
h
and
b
=
2
⇒
b
3
=
8
Also
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
b
2
+
a
b
)
=
8
+
h
−
8
=
h
⇒
(1)
∴
L
=
lim
h
→
0
1
h
(
1
a
−
1
b
)
=
−
lim
h
→
0
a
−
b
h
⋅
1
a
b
=
−
lim
h
→
0
a
3
−
b
3
h
⋅
1
a
b
(
a
2
+
b
2
+
a
b
)
=
−
lim
h
→
0
h
h
⋅
1
a
b
(
a
2
+
b
2
+
a
b
)
.....using
(
1
)
As
h
→
0
⇒
a
3
=
8
⇒
a
=
2
=
−
1
2
×
2
(
2
2
+
2
2
+
2
×
2
)
=
=
−
1
48
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9
Similar Questions
Q1
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