$$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$

$$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$$\left( {{0 \over 0}} \right)$ <br><br>Apply L Hospital Rule <br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2\left( {x - 1} \right).{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4} - 0} \over {\left( {x - 1} \right).\cos \left( {x - 1} \right) + \sin \left( {x - 1} \right)}}$$ $\left( {{0 \over 0}} \right)$ <br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^3}\cos {{\left( {x - 1} \right)}^4}} \over {\left( {x - 1} \right).\left[ {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right]}}$$ <br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4}} \over {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}}}$$ <br><br>on taking limit <br><br>= ${0 \over {1 + 1}}$ = 0