Answer:
Medium
Solution
verified
Verified by Toppr
Correct option is D)
A.
x→0
lim
x∣x∣
−
f(x)=
(−x
2
)=−0
=0
+
x∣x∣=
(+x
)=0
LHL=RHL
B.
x→
4
1
[x]=0
C.
xsin(
x
)=0×sin(
0
−1
[−1≤sin
≤1]
xs
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Answers & Comments
Answer:
Medium
Solution
verified
Verified by Toppr
Correct option is D)
A.
x→0
lim
x∣x∣
x→0
−
lim
f(x)=
x→0
−
lim
(−x
2
)=−0
2
=0
x→0
+
lim
x∣x∣=
x→0
−
lim
(+x
2
)=0
2
=0
LHL=RHL
B.
x→
4
1
−
lim
[x]=0
x→
4
1
+
lim
[x]=0
LHL=RHL
C.
x→0
−
lim
xsin(
x
1
)=0×sin(
0
−1
)=0
x→0
+
lim
xsin(
x
1
)=0×sin(
0
1
)=0
[−1≤sin
x
1
≤1]
LHL=RHL
Correct option is D)
A.
x→0
lim
x∣x∣
x→0
−
lim
f(x)=
x→0
−
lim
(−x
2
)=−0
2
=0
x→0
+
lim
x∣x∣=
x→0
−
lim
(+x
2
)=0
2
=0
LHL=RHL
B.
x→
4
1
−
lim
[x]=0
x→
4
1
+
lim
[x]=0
LHL=RHL
C.
x→0
−
lim
xsin(
x
1
)=0×sin(
0
−1
)=0
x→0
+
lim
xs