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

Let us write $$x-1=h$$ so that $$x\to 1$$ is equivalent to $$h\to 0.$$ Re-expressing everything in terms of the new variable $$h$$ gives

$$\lim_{h\to 0}\left(\dfrac{\displaystyle\int_{0}^{h^{2}}\;t\cos t^{2}\,dt}{\;h\sin h\;}\right).$$

We now approximate the numerator. For small arguments we may use the Maclaurin series $$\cos z=1-\dfrac{z^{2}}{2}+O(z^{4}).$$ Putting $$z=t^{2}$$ we obtain

$$\cos t^{2}=1-\dfrac{(t^{2})^{2}}{2}+O(t^{8})=1-\dfrac{t^{4}}{2}+O(t^{8}).$$

Hence

$$t\cos t^{2}=t-\dfrac{t^{5}}{2}+O(t^{9}).$$

We integrate this term by term from $$0$$ to $$h^{2}:$$

$$\int_{0}^{h^{2}}t\,dt=\left[\dfrac{t^{2}}{2}\right]_{0}^{h^{2}}=\dfrac{(h^{2})^{2}}{2}=\dfrac{h^{4}}{2},$$

$$\int_{0}^{h^{2}}-\dfrac{t^{5}}{2}\,dt=-\dfrac{1}{2}\left[\dfrac{t^{6}}{6}\right]_{0}^{h^{2}}=-\dfrac{(h^{2})^{6}}{12}=-\dfrac{h^{12}}{12},$$

and the next terms are of still higher order. Therefore

$$\int_{0}^{h^{2}}t\cos t^{2}\,dt=\dfrac{h^{4}}{2}+O(h^{12}).$$

Now we study the denominator. The standard Maclaurin expansion for sine is

$$\sin h=h-\dfrac{h^{3}}{6}+O(h^{5}).$$

Hence

$$h\sin h = h\left(h-\dfrac{h^{3}}{6}+O(h^{5})\right)=h^{2}-\dfrac{h^{4}}{6}+O(h^{6}).$$

Putting the two approximations together, our limit becomes

$$\lim_{h\to 0}\dfrac{\dfrac{h^{4}}{2}+O(h^{12})}{h^{2}-\dfrac{h^{4}}{6}+O(h^{6})}.$$

First we divide numerator and denominator by $$h^{2}:$$

$$=\lim_{h\to 0}\dfrac{\dfrac{h^{2}}{2}+O(h^{10})}{1-\dfrac{h^{2}}{6}+O(h^{4})}.$$

Next, using the algebraic identity $$\dfrac{1}{1-a}=1+a+O(a^{2})$$ valid for small $$a,$$ and taking $$a=\dfrac{h^{2}}{6}+O(h^{4}),$$ we have

$$\dfrac{1}{1-\dfrac{h^{2}}{6}+O(h^{4})}=1+\dfrac{h^{2}}{6}+O(h^{4}).$$

Multiplying, we get

$$\left(\dfrac{h^{2}}{2}+O(h^{10})\right)\left(1+\dfrac{h^{2}}{6}+O(h^{4})\right)=\dfrac{h^{2}}{2}+O(h^{4}).$$

As $$h\to 0,$$ the expression $$\dfrac{h^{2}}{2}+O(h^{4})$$ clearly tends to $$0.$$ Therefore the desired limit is $$0.$$

Hence, the correct answer is Option 4.