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Byju's Answer
Standard XII
Mathematics
Evaluation of a Determinant
Evaluate the ...
Question
Evaluate the following limits:
lim
x
→
π
2
−
(
t
a
n
x
)
c
o
s
x
Open in App
Solution
L
=
lim
x
→
π
2
−
(
tan
x
)
cos
x
Taking log at both sides
ln
L
=
lim
x
→
π
2
−
cos
x
ln
tan
x
As
x
→
π
2
,
tan
x
→
+
∞
So,
ln
(
tan
x
)
→
∞
. On the other hand
cos
x
coverges to
0
.
lim
x
→
π
2
−
cos
x
ln
(
tan
x
)
=
lim
x
→
π
2
−
ln
(
tan
x
)
1
cos
x
Using L'Hospitals rule,
⇒
lim
x
→
π
2
−
1
tan
(
x
)
.
1
cos
2
(
x
)
−
1
cos
2
(
x
)
.
(
−
sin
(
x
)
)
⇒
lim
x
→
π
2
−
cos
(
x
)
sin
2
(
x
)
=
0
Thus,
ln
(
L
)
=
0
∴
L
=
e
0
L
=
1
Thus,
lim
x
→
π
2
−
(
tan
x
)
cos
x
=
1
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