1. Let $I(x)=\int_0^x t\sin(10t)\,dt.$ We need to evaluate $\lim_{x\to 0}\frac{I(x)}{x}.$

2. Observe that for small $t$, $\sin(10t)\sim 10t,$ so $t\sin(10t)\sim 10t^2.$ Hence, $I(x)=\int_0^x t\sin(10t)\,dt \sim \int_0^x 10t^2\,dt=\frac{10x^3}{3}.$ Therefore, $\frac{I(x)}{x}\sim \frac{10x^3/3}{x}=\frac{10x^2}{3}\to 0 \quad (x\to 0).$

3. For a more direct verification, apply the standard result: If $f$ is continuous at $0$, then $\lim_{x\to 0}\frac{1}{x}\int_0^x f(t)\,dt=f(0).$ Here, $f(t)=t\sin(10t),$ which is continuous at $t=0$, and $f(0)=0\cdot \sin 0=0.$ Thus, $\lim_{x\to 0} \frac{\int_0^x t\sin(10t)\,dt}{x}=0.$

4. Check the options:

• A: $-\frac15$ ❌
• B: $-\frac1{10}$ ❌
• C: $0$ ✅
• D: $\frac1{10}$ ❌

So the correct option is C.