# Use Taylor's theorem to evaluate the following limits. lim_{xrightarrow0}frac{xsin(x)-x^2}{cos(x)-1+frac{x^2}{2}}

Question
Limits and continuity
Use Taylor's theorem to evaluate the following limits. $$\lim_{x\rightarrow0}\frac{x\sin(x)-x^2}{\cos(x)-1+\frac{x^2}{2}}$$

2020-10-27
Evaluate limit using Taylor’s theorem.
Given:
$$\lim_{x\rightarrow0}\frac{x\sin(x)-x^2}{\cos(x)-1+\frac{x^2}{2}}$$
Taylor series expansion of trigonometric functions,
$$\sin x=\frac{x-x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$, $$-\infty Substitute the series, \(\lim_{x\rightarrow0}\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}$$
$$\lim_{x\rightarrow0}\frac{(x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+...) -x^2}{\frac{1-x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)-1+\frac{x^2}{2})}$$
Cancelling the term,
$$\lim_{x\rightarrow0}\frac{\frac{-x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+...}{\frac{x^4}{4!}-\frac{x^6}{6!}+...}$$
Divide $$x^4$$ in both numerator and denominator,
$$\lim_{x\rightarrow0}\frac{\frac{-1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+...}{\frac{1}{4!}-\frac{x^2}{6!}+..}$$
Apply limit, x tends to 0.
$$\Rightarrow\frac{\frac{-1}{3!}}{\frac{1}{4!}}$$
$$\Rightarrow\frac{-4!}{3!}$$
$$\Rightarrow\frac{4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}$$
$$\Rightarrow-4$$

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