Answer
= $1$
Work Step by Step
$\mathop {\lim }\limits_{x \to 0 }$ $\frac{x(sin{x^{2}})}{tan^{3}x}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{x(cos{x^{2}})(2x)+sinx^{2}(1)}{3tan^{2}x(sec^{2}x)}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{2x^{2}(cos{x^{2}})+sinx^{2}}{3tan^{2}x(1+tan^{2}x)}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{2x^{2}(cos{x^{2}})+sinx^{2}}{3tan^{2}x+3tan^{4}x}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{2x^{2}(-sin{x^{2}})(2x)+(cos{x^{2}})(4x)+cosx^{2}(2x)}{6tan(x)sec^{2}x+12tan^{3}x(sec^{2}x)}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{-4x^{3}sin{x^{2}}+6x(cos{x^{2}})}{6tan{x}(1+tan^{2}x)+12tan^{3}x(1+tan^{2}x)}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{-4x^{3}sin{x^{2}}+6x(cos{x^{2}})}{6tan{x}+6tan^{2}x+12tan^{3}x+12tan^{5}x}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{-4x^{3}sin{x^{2}}+6x(cos{x^{2}})}{6tan{x}+6tan^{2}x+12tan^{3}x+12tan^{5}x}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{-4x^{3}cosx^{2}(2x)+sinx^{2}(-12x^{2})+6x(-sin{x^{2}})(2x)+cosx^{2}(6)}{{6sec^{2}x}+6(2)(tanx)(sec^{2}x)+12(3)tan^{2}x(sec^{2}x)+12(5)tan^{4}x(sec^{2}x)}$
= $\mathop {\lim }\limits_{x \to 0 }$ $\frac{(-8x^{4}+6)cosx^{2}-24x^{2}sinx^{2}}{{6sec^{2}x}+12(tanx)(sec^{2}x)+36tan^{2}x(sec^{2}x)+60tan^{4}x(sec^{2}x)}$
=$\frac{6}{6}$
= $1$
