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Question Number 113786 by bemath last updated on 15/Sep/20

Answered by bobhans last updated on 15/Sep/20

$$recall\tan 6x=6x+\frac{{\left(6x\right)}^3}{3}+\frac{2{\left(6x\right)}^5}{15}+...$$$$\cos \left(\frac{4}{x}\right)=1-\frac{{\left(\frac{4}{x}\right)}^2}{2!}+\frac{{\left(\frac{4}{x}\right)}^4}{4!}-...$$$$\underset{x\to 0}{\lim }\frac{2(6x+\frac{6^3{x}^3}{3}+\frac{2.6^5{x}^5}{15}+...)}{1-(1-\frac{8}{x^2}+\frac{8}{3x^4}-...)}=$$$$\underset{x\to 0}{\lim }\frac{2(6x+\frac{6^3{x}^3}{3}+\frac{2.6^5{x}^5}{15}+...)}{\frac{1}{x^2}(8+\frac{8}{3x^2}-...)}=$$$$\underset{x\to 0}{\lim }\frac{2x^2(6x+\frac{6^3{x}^3}{3}+\frac{2.6^5{x}^5}{15}+...)}{8+\frac{8}{3x^2}-...}=0$$

Answered by bemath last updated on 15/Sep/20

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