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Question Number 185510 by mathlove last updated on 23/Jan/23

Answered by mahdipoor last updated on 23/Jan/23

$$6x-8sin\left(x\right)+sin\left(2x\right)=$$$$6x-8(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....)+$$$$(2x-\frac{8x^3}{3!}+\frac{32x^5}{5!}-\frac{128x^7}{7!}+...)=$$$$(6-8+2)x+(\frac{8}{3!}-\frac{8}{3!})x^3+(\frac{-8}{5!}+\frac{32}{5!})x^5+{ax}^7+....$$$$=\frac{24}{5!}{x}^5+{ax}^7+...$$$$\underset{x\to 0}{\lim }\frac{6x-8sinx+sin2x}{x^5}=\underset{x\to 0}{\lim }(\frac{24}{5!}+{ax}^2+...)=$$$$\frac{24}{5!}=\frac{1}{5}$$

Commented by MJS_new last updated on 23/Jan/23

$$\mathrm{but}\mathrm{it}\mathrm{says}‘‘without{\mathrm{series}}^{″}...$$

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