Question

could u pls solve this lim x→π/2 (sin x)^tan x​

Answer 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.

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Answer 2

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

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