#### To determine

**To find:** The limit of the expression limt→0+[A(t)B(t)].

**Solution:**0.

**Given: **The expression is limt→0+[A(t)B(t)]

**Formulae used:** L’Hospital rule

For a given problem limx→cf(x)g(x), if the limits of f(x) and g(x) are 00 or ±∞∞ and if the limit of their derivatives limx→cf'(x)g'(x) is an extended real umber, then The L’ Hospital rule limx→0f(x)g(x)=limx→0f′(x)g′(x).

#### Explanation

The area under the curve y=sin(x2) from x=0 to x=t is

A(x)=∫0tsin(x2) dxA′(t)=sin(t2)

And the area of the triangle with vertices *O, P, *and (t,0) is

B(t)=12×t×sin(t2)=12tsin(t2)B′(t)=12[sin(t2)+2t2cos(t2)]

Since limt→0+[A(t)B(t)] is of 00 form we use l'Hospital's rule. Thuslimt→0+[A(t)B(t)]=limt→0+[A′(t)B′(t)]=limt→0+[2sin(t2)sin(t2)+2t2cos(t2)]=0

**Conclusion:**Hence the limit of the expression limt→0+[A(t)B(t)] is 0.