# Use Taylor's theorem to evaluate the following limits. lim_{xrightarrow0}frac{3sin^2(x)+2sin^4(x)}{3xtan(x)}

Question
Limits and continuity
Use Taylor's theorem to evaluate the following limits. $$\lim_{x\rightarrow0}\frac{3\sin^2(x)+2\sin^4(x)}{3x\tan(x)}$$

2021-02-24
Evaluate limit using Taylor’s theorem.
Given:
$$\lim_{x\rightarrow0}\frac{3\sin^2(x)+2\sin^4(x)}{3x\tan(x)}$$
Taylor series expansion of trigonometric functions,
$$\sin^2x=x^2-\frac{x^4}{3}+\frac{2x^6}{45}-\frac{x^8}{315}+...$$
$$\sin^4x=x^4-\frac{2x^6}{3}+\frac{x^8}{5}-\frac{34x^{10}}{945}+...$$
$$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}+...$$
Substitute the series,
$$\lim_{x\rightarrow0}\frac{3(x^2-\frac{x^4}{3}+\frac{2x^6}{45}-\frac{x^8}{315}+...)+2(x^4-\frac{2x^6}{3}+\frac{x^8}{5}-\frac{34x^{10}}{945}+...)}{3x(x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}+...)}$$
$$\lim_{x\rightarrow0}\frac{(3x^2-\frac{3x^4}{3}+\frac{3(2x^6)}{45}-\frac{3x^8}{315}+...)+(2x^4-\frac{2(2x^6)}{3}+\frac{2x^8}{5}-\frac{2(34x^{10})}{945}+...)}{(3x^2+\frac{3x^4}{3}+\frac{3\cdot2x^6}{15}+\frac{3\cdot17x^8}{315}+\frac{3\cdot62x^{10}}{2835}+...))}$$
Divide $$x^2$$ in both numerator and denominator,
$$\lim_{x->0}\frac{(3-\frac{3x^2}{3}+\frac{3(2x^4)}{45}-\frac{3(x^6)}{315}+...)+(2x^2-\frac{2(2x^4)}{3}+\frac{2x^6}{5}-\frac{2(34x^8)}{945}+...)}{(3+\frac{3x^2}{3}+\frac{(3\cdot2x^4)}{15}+\frac{(3\cdot17x^6)}{315}+\frac{(3\cdot62x^8)}{2835}+...)}$$
Apply limit, x tends to 0
$$=3/3$$
$$\Rightarrow1$$
Result: 1

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