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Question Number 128135 by help last updated on 04/Jan/21

Commented by liberty last updated on 04/Jan/21

$$\underset{x\to 0}{\lim }{\left(\frac{3\sin 5\mathrm{x}}{\mathrm{x}}\right)}^{\underset{x\to 0}{\lim }\left(\frac{1-\cos 4\mathrm{x}}{{\mathrm{x}}^2}\right)}={15}^{\underset{x\to 0}{\lim }\left(\frac{1-(1-\frac{{16\mathrm{x}}^2}{2})}{{\mathrm{x}}^2}\right)}$$$$={15}^{\underset{x\to 0}{\lim }\left(\frac{{8\mathrm{x}}^2}{{\mathrm{x}}^2}\right)}={15}^8.$$

Answered by Dwaipayan Shikari last updated on 04/Jan/21

$$\underset{x\to 0}{\lim }\left(\frac{3sin5x}{x}\right)\left(\frac{1-cos4x}{x^2}\right)$$$$=15.\left(\frac{2{sin}^{2}2x}{x^2}\right)=30.\left(\frac{4x^2}{x^2}\right)=120$$

Answered by Dwaipayan Shikari last updated on 04/Jan/21

$$\underset{x\to 0}{\lim }{\left(\frac{3sin5x}{x}\right)}^{\left(\frac{1-cos4x}{x^2}\right)}$$$${\left(15\right)}^{\frac{2{sin}^{2}2x}{x^2}}={15}^8$$

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