Find the limit of (1+sinx)^(1/x) when x approaches 0

Evaluate limit as x approaches 0 of 1/(sin(x))-1/x
lim
x
→
0

1
sin
(
x
)
−
1
x
lim
x
→
0
⁡
1
sin
(
x
)
-
1
x
Convert from
1
sin
(
x
)
1
sin
(
x
)
to
csc
(
x
)
csc
(
x
)
.
lim
x
→
0

csc
(
x
)
−
1
x
lim
x
→
0
⁡
csc
(
x
)
-
1
x
Consider the left sided limit.
lim
x
→
0
−

csc
(
x
)
−
1
x
lim
x
→
0
-
⁡
csc
(
x
)
-
1
x
Make a table to show the behavior of the function
csc
(
x
)
−
1
x
csc
(
x
)
-
1
x
as
x
x
approaches
0
0
from the left.
x
csc
(
x
)
−
1
x
−
0.1
−
0.01668613
−
0.01
−
0.00166668
−
0.001
−
0.0001
¯
6
−
0.0001
−
0.00001
¯
6
xcsc
(
x
)
-
1
x
-0.1-0.01668613-0.01-0.00166668-0.001--0.0001-
As the
x
x
values approach
0
0
, the function values approach
0
0
. Thus, the limit of
csc
(
x
)
−
1
x
csc
(
x
)
-
1
x
as
x
x
approaches
0
0
from the left is
0
0
.
0
0
Consider the right sided limit.
lim
x
→
0
+

csc
(
x
)
−
1
x
lim
x
→
0
+
⁡
csc
(
x
)
-
1
x
Make a table to show the behavior of the function
csc
(
x
)
−
1
x
csc
(
x
)
-
1
x
as
x
x
approaches
0
0
from the right.
x
csc
(
x
)
−
1
x
0.1
0.01668613
0.01
0.00166668
0.001
0.0001
¯
6
0.0001
0.00001
¯
6
xcsc
(
x
)
-
1
x
0.10.016686130.010.001666680.001 0.0001
As the
x
x
values approach
0
0
, the function values approach
0
0
. Thus, the limit of
csc
(
x
)
−
1
x
csc
(
x
)
-
1
x
as
x
x
approaches
0
0
from the right is
0
0
.
0

Ithink the right answer could be 50%