1. We need to evaluate

$$L=\lim_{x\to 1}\frac{\int_0^{(x-1)^2} t\cos(t^2)\,dt}{(x-1)\sin(x-1)}.$$

2. First simplify the numerator using substitution.

Let

$$u=t^2 \implies d\nu=2t\,dt \implies t\,dt=\frac{d\nu}{2}.$$

When $t=0$, $\nu=0$; when $t=(x-1)^2$,

$$\nu=((x-1)^2)^2=(x-1)^4.$$

So,

$$\int_0^{(x-1)^2} t\cos(t^2)\,dt =\frac12\int_0^{(x-1)^4} \cos \nu\,d\nu =\frac12\sin\big((x-1)^4\big).$$

Thus,

$$L=\lim_{x\to 1} \frac{\frac12\sin((x-1)^4)}{(x-1)\sin(x-1)} =\frac12\lim_{x\to 1}\frac{\sin((x-1)^4)}{(x-1)\sin(x-1)}.$$

3. Put $h=x-1$. Then as $x\to 1$, $h\to 0$. The limit becomes

$$L=\frac12\lim_{h\to 0}\frac{\sin(h^4)}{h\sin h}.$$

4. Use the standard small-angle fact:

$$\sin u \sim u \quad (u\to 0).$$

Hence,

$$\sin(h^4)\sim h^4, \qquad \sin h\sim h.$$

Therefore,

$$\frac{\sin(h^4)}{h\sin h} \sim \frac{h^4}{h\cdot h}=h^2.$$

As $h\to 0$,

$$h^2\to 0.$$

So,

$$L=\frac12\cdot 0=0.$$

5. Therefore the correct option is:

$$\boxed{\text{A: is equal to }0}$$

6. Verification with stored answer:

• Derived answer: A
• Stored correct answer: A
• They agree.