could u pls solve this lim x→π/2 (sin x)^tan x
Answers
Answered by
1
Answer:
HOPE THIS HELPS-
lim (x → π/2) (sinx)^(tanx)
= lim (x → π/2) e^[(tanx) ln (sinx)]
= e^ [lim (x → π/2) (tanx) ln (sinx)] ... (1)
lim (x → π/2) (tanx) ln (sinx)
= lim (x → π/2) [ln(sinx) / cotx]
Using L'Hospital'stheorem,
= lim (x → π/2) [- cotx / cosec^2 x]
= 0
Plugging in ( 1 ),
required limit = e^0 = 1
=>Answer is 1.
MARK AS BRAINLIEST... :)
Answered by
0
Answer:
Step-by-step explanation:
Lim(x→π/2) e^{tanx.logsinx}
Lim(x→π/2)e^{logsinx}/cotx
e^[Lim(x→π/2){logsinx}/{cotx}]
now check the form of limit . put x = π/2
so,logsinπ/2/cotπ/2 = 0/0
hence, now we can use L - Hospital rule .
so, e^[Lim(x→π/2){cosx/sinx}/{-cosec²x}]
= e^[Lim(x→π/2){cosx/-sinx)}
now, put x = π/2
we get ,
= e^{0/1} = e^0 = 1
MARK IT THE BRAINLIEST
Similar questions