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Question Number 173498 by blackmamba last updated on 12/Jul/22

$$\underset{x\to 0}{\lim }\frac{8\cot \left(x\right)+9\sin \left(\frac{1}{x}\right)}{12\csc \left(x\right)-4\sin \left(\frac{1}{x}\right)}=?$$

Commented by blackmamba last updated on 13/Jul/22

$$=\underset{x\to 0}{\lim }\frac{8\cos x\left(\frac{x}{\sin x}\right)+9x\sin \left(\frac{1}{x}\right)}{\frac{12x}{\sin x}-4x\sin \left(\frac{1}{x}\right)}$$$$=\frac{8+0}{12-0}=\frac{8}{12}=\frac{2}{3}$$

Answered by aleks041103 last updated on 12/Jul/22

$$L=\left(\underset{x\to 0}{lim}\frac{8cot\left(x\right)}{12csc\left(x\right)}\right)\left(\underset{x\to 0}{lim}\frac{1+\frac{9}{8}tan\left(x\right)sin\left(1/x\right)}{1-\frac{1}{3}sin\left(x\right)sin\left(1/x\right)}\right)=$$$$=\frac{2}{3}cos\left(0\right).\frac{1+\frac{9}{8}.0.\left(sthfinite\right)}{1-\frac{1}{3}.0.\left(sthfinite\right)}=\frac{2}{3}$$$$\Rightarrow Ans.=\frac{2}{3}$$

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