$$\lim_{x \to 0} \frac{\int_0^x t \sin(10t) dt}{x}$$ is equal to

We have to evaluate the limit

$$\lim_{x \to 0} \dfrac{\displaystyle \int_{0}^{x} t \sin(10t)\,dt}{x}.$$

At first glance both the numerator and the denominator vanish when $$x = 0$$, giving an indeterminate form $$\dfrac{0}{0}$$. A convenient way to handle such a form is L’Hospital’s Rule.

L’Hospital’s Rule: If $$\displaystyle\lim_{x\to a} \dfrac{f(x)}{g(x)} = \dfrac{0}{0}$$ or $$\dfrac{\pm\infty}{\pm\infty}$$ and the derivatives exist near $$a$$, then

$$\lim_{x\to a} \dfrac{f(x)}{g(x)} = \lim_{x\to a} \dfrac{f'(x)}{g'(x)}.$$

In our problem we identify

$$f(x)=\int_{0}^{x} t\sin(10t)\,dt, \qquad g(x)=x.$$

We differentiate each with respect to $$x$$.

First, by the Fundamental Theorem of Calculus, the derivative of the integral is simply the integrand evaluated at the upper limit:

$$f'(x)=x\sin(10x).$$

Next, the derivative of $$g(x)=x$$ is

$$g'(x)=1.$$

Applying L’Hospital’s Rule, the original limit equals the new limit

$$\lim_{x\to 0} \dfrac{f'(x)}{g'(x)}=\lim_{x\to 0} \dfrac{x\sin(10x)}{1}= \lim_{x\to 0} x\sin(10x).$$

Now we examine $$x\sin(10x)$$ as $$x\to 0$$. We recall the standard small-angle fact:

$$\sin\theta \approx \theta \text{ when } \theta\to 0.$$

Here $$\theta = 10x$$, so $$\sin(10x) \approx 10x$$ for very small $$x$$. Substituting this approximation we get

$$x\sin(10x) \approx x\,(10x)=10x^{2}.$$

Clearly $$10x^{2}\to 0$$ as $$x\to 0$$, and therefore

$$\lim_{x\to 0} x\sin(10x)=0.$$

Hence the required limit is $$0$$.

Hence, the correct answer is Option A.