pls solve by using l hopital
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Answer:
lim
x→
2
π
(secx−tanx)
⇒lim
x→
2
π
[
cosx
1
−
cosx
sinx
]
⇒lim
x→
2
π
[
cosx
1−sinx
]
⇒lim
x→π/2
cos
2
2
x
−sin
2
2
x
(cosx
2
x
−sin
2
x
)
2
{∴1=sin
2
dfracx2+cos
2
x/2,sinx=2sinx/2cosx/2,cosx=cos
2
2
x
−sin
2
2
x
}
⇒lim
x→
2
π
(cosx/2−sinx/2)(cosx/2+sinx/2)
(cosx/2−sinx/2)
2
⇒lim
x→
2
π
cosx/2+sinx/2
cosx/2−sinx/2
⇒0
please mark me as brainly answer
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