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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(x)+sin(2x)= 6x−8(x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!))+....)+ (2x−((8x^3 )/(3!))+((32x^5 )/(5!))−((128x^7 )/(7!))+...)= (6−8+2)x+((8/(3!))−(8/(3!)))x^3 +(((−8)/(5!))+((32)/(5!)))x^5 +ax^7 +.... =((24)/(5!))x^5 +ax^7 +... lim_(x→0) ((6x−8sinx+sin2x)/x^5 )=lim_(x→0) (((24)/(5!))+ax^2 +...)= ((24)/(5!))=(1/5)

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

Commented by MJS_new last updated on 23/Jan/23

but it says “without series”...

$$\mathrm{but}\:\mathrm{it}\:\mathrm{says}\:``{without}\:\mathrm{series}''... \\ $$

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