Question

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Answer 1

Answer:

1/2

Step-by-step explanation:

If we apply lim x_π/2

x=π/2

then lim x_π/2 =1-1/0-0 =1/0=undetermined

sin(π/2)=0

(π/2-π/2)=0

so Then we can consider

x=π/2-h become lim h_0

lim h_0=(1-sin(π/2-h))/((π/2-(π/2-h)cot(π/2-h))

lim h_0=(1-cosh)/htanh

=(2sin^2h/2)/htan

(sinh/2)/(h/2)=1

(sin h^2/2)/(h/2)^2)=1

then

lim h_0= (2sin^2h/2)/(h^2/4 ×tanh×4/h)

lim h_0=(2×1)/tanh×4/h

lim h_0=h/2×tanh=1/2

h/tanh=1

Answer 2

Answer:

1/2

Step-by-step explanation:

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