#### To determine

**To find:** The value of limθ→0+A(θ)B(θ) .

#### Answer

The value of limθ→0+A(θ)B(θ)=0.

#### Explanation

**Given:**

The area of the semicircle is A(θ) and the area of the triangle is B(θ).

**Formula used:**

The area of the semicircle is 12πr2 (1)

The area of the triangle is 12bh (2)

**Calculation:**

The given picture is shown in the below Figure 1.

From Figure 1, the base of the triangle (*PQ*) is same as the diameter of semicircle (2*r*).

Calculate *r* and *h*.

From the triangle *PSR*,

sinθ2=opposite sidehypotenuse=r10r=10sinθ2

And

cosθ2=adjacenthypotenuse=h10h=10cosθ2

Substitute b=2r in equation (2),

B(θ)=12(2r)h=rh

Thus, the area of the triangle is B(θ)=rh.

From equation (1), the area if the semicircle is A(θ)=12πr2.

Consider limθ→0+A(θ)B(θ).

Substitute A(θ)=12πr2 and B(θ)=rh,

limθ→0+A(θ)B(θ)=limθ→0+12πr2rh=limθ→0+12πrh=π2⋅limθ→0+rh

Substitute r=10sinθ2 and h=10cosθ2,

limθ→0+A(θ)B(θ)=π2⋅limθ→0+10sinθ210cosθ2=π2⋅limθ→0+(tanθ2 ) (Qsinxcosx=tanx)=π2⋅tan(0)=0

Therefore, the value of limθ→0+A(θ)B(θ)=0.