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Test Index
Limits, Continuity and Differentiability Part 1
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Section:
Mathematics
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© examsnet.com
Question : 3
Total: 100
lim
x
→
0
x
(
e
(
√
1
+
x
2
+
x
4
−
1
)
∕
x
−
1
)
√
1
+
x
2
+
x
4
−
1
[5 Sep 2020 Shift 2]
does not exist.
is equal to
√
e
.
is equal to 0 .
is equal to 1.
Validate
Solution:
lim
x
→
0
x
(
e
(
√
1
+
x
2
+
x
4
−
1
)
∕
x
−
1
)
√
1
+
x
2
+
x
4
−
1
∵
lim
x
→
0
√
1
+
x
2
+
x
4
−
1
x
(
0
0
from
)
lim
x
→
0
(
1
+
x
2
+
x
4
)
−
1
x
(
√
1
+
x
2
+
x
4
+
1
lim
x
→
0
x
(
1
+
x
2
)
(
√
1
+
x
2
+
x
4
+
1
)
=
0
So
lim
x
→
0
x
(
e
(
√
1
+
x
2
+
x
4
−
1
x
)
−
1
)
√
1
+
x
2
+
x
4
−
1
‌
(
0
0
from
)
lim
x
→
0
e
√
1
+
x
2
+
x
4
−
1
x
−
1
(
√
1
+
x
2
+
x
4
−
1
x
)
=
1
© examsnet.com
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