Limit of Integral: x $\rightarrow$ 0, $\int{\sin{t^2}}dt$ 0 to 3

[ultrazyn]

limit as x approaches 0 of [(1/x^3) * integral from 0 to 3 of (sin(t^2))dt]

is the way to solve this using mean value theorem for integrals?

$$\lim_{x\rightarrow0}\frac{1}{x^3}\int{\sin{t^2}}dt$$ the integral is from 0 to 3
not sure if i did the latex right

 

[Orion1]

limit as x approaches 0 of [(1/x^3) * integral from 0 to 3 of (sin(t^2))dt]

$$\lim_{x \rightarrow 0} \frac{1}{x^3} \int_0^3 \sin{t^2} dt$$

 

[HallsofIvy]

If the integral is, in fact, from 0 to 3, then it is just a constant. This is asking for
$$\lim_{x \rightarrow 0} \frac{c}{x^2}$$
where c is a non-zero constant. Looks pretty easy to me.

 

[mathman]

The problem makes more sense (as a problem) if the upper limit of the integral is x, not 3. In that case, the answer is 1/3, assuming that the divisor is x³, not x².