1,655,000 maybe I not entirely sure sorry
Answer:
the answer to this question is the number 8
Answer:
no
Step-by-step explanation:
<span>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.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
The correct answer is C. 78
