## Task $9 / 75$

Determine, without calculating the integral, the limit

$$
\lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} t \sin t d t
$$
Is $F(t)=\int t \cdot \sin t d t$ an antiderivative of $t \cdot \sin t$, then the limit takes the form

$$
\lim _{x \rightarrow 0} \frac{F(x)-F(0)}{x-0}
$$

One recognizes that the limit of the difference quotient of $F(x)$ at the point $x=0$ is to be formed; this is the derivative of the function $F(x)$ at the point $x=0$, i.e., $\left.F^{\prime}(x)\right|_{x=0}$. Since $F(x)$ is an antiderivative of $x \sin x$, it follows that $F^{\prime}(x) = x \sin x$. Therefore, the desired limit is

$$
\left.x \sin x\right|_{x=0}=0
$$

Note: This method can be applied to any limit of the form

$$
\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(t) d t}{x}
$$

even if the integral is not evaluable in elementary terms.