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Question Number 91954 by john santu last updated on 04/May/20

$$\underset{x\to 0^{+}}{\lim }\frac{\ln \left(\sin 2\mathrm{x}\right)}{\ln \left(\sin 4\mathrm{x}\right)}$$

Commented by jagoll last updated on 04/May/20

$${\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented by jagoll last updated on 04/May/20

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Commented by abdomathmax last updated on 04/May/20

$$hospitaltheorem$$$${lim}_{x\to 0^{+}}\frac{ln\left(sin\left(2x\right)\right)}{ln\left(sin\left(4x\right)\right)}={lim}_{x\to 0^{+}}\frac{\frac{2cos\left(2x\right)}{sin\left(2x\right)}}{\frac{4cos\left(4x\right)}{sin\left(4x\right)}}$$$$={lim}_{x\to 0^{+}}\frac{2cos\left(2x\right)}{sin\left(2x\right)}\times \frac{sin\left(4x\right)}{4cos\left(4x\right)}$$$$={lim}_{x\to 0^{+}}\frac{1}{2}\frac{cos\left(2x\right)}{cos\left(4x\right)}\times \frac{sin\left(4x\right)}{sin\left(2x\right)}$$$$=\frac{1}{2}\times 1\times 2=1$$

Commented by jagoll last updated on 04/May/20

yes ������

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