Will you use L'hopital? (1)

Relevant wiki: L'Hopital's Rule - Intermediate

$\begin{aligned} \mathscr L & = \lim_{x \to 0} \frac {1-\cos x +2\sin x -\sin ^3 x -x^2+3x^4}{\tan^3 x - 6\sin^2 x +x-5x^3} \quad \quad \small \color{#3D99F6}{\text{Since this is a } 0 / 0 \text{ case, we can apply L'Hôpital's rule.}} \\ & = \lim_{x \to 0} \frac {\sin x +2\cos x -3\sin^2 x \cos x -2x+12x^3}{3\tan^2 x\sec^2 x - 12\sin x\cos x +1-15x^2} \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down w.r.t. } x.} \\ & = \boxed{2} \end{aligned}$

Using the maclaurin series of $\sin{x}$ $\tan{x}$ and $\cos{x}$ . The given limit transforms to $\lim _{ x\rightarrow 0 }{ \frac { 1-\left( 1-\frac { { x }^{ 2 } }{ 2! } +.... \right) +2\left( x-\frac { { x }^{ 3 } }{ 3! } +..... \right) -{ \left( x-\frac { { x }^{ 3 } }{ 3! } +....... \right) }^{ 3 }-{ x }^{ 2 }+3{ x }^{ 4 } }{ { {\left( x+\frac { { x }^{ 3 } }{ 3 } +....... \right)}^{3} -6{ \left( x-\frac { { x }^{ 3 } }{ 3! } +..... \right) }^{ 2 }+x-5{ x }^{ 3 } } } }$ Now taking $x$ common from both ${N}^{r}$ and ${D}^{r}$ . We get $\lim _{ x\rightarrow 0 }{ \frac { \frac { x }{ 2! } +2\left( 1-\frac { { x }^{ 2 } }{ 3! } +..... \right) -{ { x }^{ 2 }\left( 1-\frac { { x }^{ 2 } }{ 3! } +....... \right) }^{ 3 }-{ x }+3{ x }^{ 3 } }{ { { x }^{ 2 }{ \left( 1-\frac { { x }^{ 2 } }{ 3 } +....... \right) }^{ 3 }-6{ x\left( 1-\frac { { x }^{ 2 } }{ 3! } +..... \right) }^{ 2 }+1-5{ x }^{ 2 } } } } =2$

As x goes to 0, sin x~ x and cos x~ 1-x^2, substitute wherever suitable and the result follows

Divide numerator and denominator by x and solve.