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Byju's Answer
Standard VIII
Mathematics
Identities of Mathematical Operations
Evaluate: l...
Question
Evaluate:
lim
x
→
1
√
x
+
8
−
√
8
x
+
1
√
5
−
x
−
√
7
x
−
3
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Solution
Method of rationalising both numerators and denominators,
=
lim
x
→
1
(
√
x
+
8
−
√
8
x
+
1
)
(
√
5
−
x
+
√
7
x
−
3
)
(
√
5
−
x
−
√
7
x
−
3
)
(
√
5
−
x
+
√
7
x
−
3
)
=
lim
x
→
1
(
√
(
x
+
8
)
−
√
8
x
+
1
)
(
√
5
−
x
+
√
7
x
−
3
)
5
−
x
−
7
x
+
3
=
lim
x
→
1
(
√
x
+
8
−
√
8
x
+
1
)
(
√
5
−
x
+
√
7
x
−
3
)
−
8
x
+
8
×
√
x
+
8
+
√
8
x
+
1
√
x
+
8
+
√
8
x
+
1
=
lim
x
→
1
(
√
x
+
8
−
√
8
x
+
1
)
(
√
5
−
x
+
√
7
x
−
3
)
(
√
x
+
8
+
√
8
x
+
1
)
−
8
(
x
−
1
)
(
(
√
x
+
8
+
√
8
x
+
1
)
)
=
lim
x
→
1
(
x
+
8
−
8
x
−
1
)
(
√
5
−
x
+
√
7
x
−
3
)
−
8
(
x
−
1
)
(
(
√
x
+
8
+
√
8
x
+
1
)
)
=
lim
x
→
1
(
−
7
x
+
7
)
(
√
5
−
x
+
√
7
x
−
3
)
−
8
(
x
−
1
)
(
√
x
+
8
+
√
8
x
+
1
)
=
lim
x
→
1
−
7
(
x
−
1
)
(
√
5
−
x
+
√
7
x
−
3
)
−
8
(
x
−
1
)
(
√
x
+
8
+
√
8
x
+
1
)
=
lim
x
→
1
7
(
√
5
−
x
+
√
7
x
−
3
)
8
(
√
x
+
8
+
√
8
x
+
1
)
=
7
(
√
5
−
1
+
√
7
×
1
−
3
)
8
(
√
1
+
8
+
√
8
×
1
+
1
)
=
7
(
√
4
+
√
4
)
8
(
√
9
+
√
9
)
=
14
√
4
16
√
9
=
14
×
2
16
×
3
=
7
4
×
3
=
7
12
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