Use Taylor's theorem to evaluate the following limits. lim_{x->0}({xsin(x)-x^2)/(cos(x)-1+(x^2)/(2))

Use Taylor's theorem to evaluate the following limits. $\underset{x\to 0}{lim}\frac{x\mathrm{sin}\left(x\right)-{x}^{2}}{\mathrm{cos}\left(x\right)-1+\frac{{x}^{2}}{2}}$

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Evaluate limit using Taylor’s theorem.
Given:
$\underset{x\to 0}{lim}\frac{x\mathrm{sin}\left(x\right)-{x}^{2}}{\mathrm{cos}\left(x\right)-1+\frac{{x}^{2}}{2}}$
Taylor series expansion of trigonometric functions,
$\mathrm{sin}x=\frac{x-{x}^{3}}{3!}+\frac{{x}^{5}}{5!}-\frac{{x}^{7}}{7!}+...$, $-\mathrm{\infty }$
Substitute the series,
$\underset{x\to 0}{lim}\frac{x\frac{x-{x}^{3}}{3!}+\frac{{x}^{5}}{5!}-\frac{{x}^{7}}{7!}+...}{1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+...}-1+\frac{{x}^{2}}{2}$
$\underset{x\to 0}{lim}\frac{\left({x}^{2}-\frac{{x}^{4}}{3!}+\frac{{x}^{6}}{5!}-\frac{{x}^{8}}{7!}+...\right)-{x}^{2}}{\frac{1-{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+...\right)-1+\frac{{x}^{2}}{2}\right)}$
Cancelling the term,
$\underset{x\to 0}{lim}\frac{\frac{-{x}^{4}}{3!}+\frac{{x}^{6}}{5!}-\frac{{x}^{8}}{7!}+...}{\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+...}$
Divide ${x}^{4}$ in both numerator and denominator,
$\underset{x\to 0}{lim}\frac{\frac{-1}{3!}+\frac{{x}^{2}}{5!}-\frac{{x}^{4}}{7!}+...}{\frac{1}{4!}-\frac{{x}^{2}}{6!}+..}$
Apply limit, x tends to 0.
$⇒\frac{\frac{-1}{3!}}{\frac{1}{4!}}$
$⇒\frac{-4!}{3!}$
$⇒\frac{4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}$
$⇒-4$