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Question Number 111103 by bemath last updated on 02/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\underset{x\to 0}{\lim }\frac{\ln \left(\sin 3\mathrm{x}\right)}{\ln \left(\sin 8\mathrm{x}\right)}?$$$$\left[\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{Hopital}\right]$$$$$$

Answered by Lordose last updated on 02/Sep/20

$$\mathbf{what}\mathbf{about}\mathbf{series}...$$

Commented by bemath last updated on 02/Sep/20

$$\mathrm{yes}..$$

Answered by bobhans last updated on 02/Sep/20

$$\mathrm{If}\mathrm{by}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 0}{\lim }\frac{3\cot 3\mathrm{x}}{8\cot 8\mathrm{x}}=\underset{x\to 0}{\lim }\frac{3\tan 8\mathrm{x}}{8\tan 3\mathrm{x}}=1$$

Answered by Dwaipayan Shikari last updated on 02/Sep/20

$$\underset{x\to 0}{\lim }\frac{log\left(sin3x\right)}{log\left(sin8x\right)}=\frac{log\left(3x\right)}{3x}.\frac{8x}{log\left(8x\right)}=\frac{3x-\frac{9x^2}{2}}{3x}.\frac{8x}{8x-\frac{64x^2}{2}}=1$$$$\frac{9x^2}{2}\to 0,32x^2\to 0$$$$sinx\to x$$

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