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Question Number 130422 by john_santu last updated on 25/Jan/21

$${\lim }_{\mathrm{x}\to 0}\left(\frac{\ln (1+\sin 2x^5)}{\tan {}^{3}3x(5^{x^2}-1)}\right)=?$$

Answered by liberty last updated on 25/Jan/21

$$\underset{x\to 0}{\lim }\left(\frac{\ln (1+\sin 2x^5)}{\sin 2x^5}\right).\underset{x\to 0}{\lim }\frac{\sin 2x^5}{{\left(\tan 3x\right)}^3(5^{x^2}-1)}=$$$$1\times \underset{x\to 0}{\lim }\frac{\sin 2x^5}{2x^5}\times \underset{x\to 0}{\lim }\frac{2x^5}{{\left(\tan 3x\right)}^3(5^{x^2}-1)}=$$$$1\times 1\times \underset{x\to 0}{\lim }\frac{2x^2}{{\left(\frac{\tan 3x}{x}\right)}^3(5^{x^2}-1)}=$$$$1\times 1\times \frac{2}{27}\times \underset{x\to 0}{\lim }\left(\frac{x^2}{5^{x^2}-1}\right)=$$$$let{x}^2={\ell }^2\Rightarrow 1\times 1\times \frac{2}{27}\times \frac{1}{\underset{\ell \to 0}{\lim }\left(\frac{5^{\ell }-1}{\ell }\right)}=$$$$1\times 1\times \frac{2}{27}\times \frac{1}{\ln \left(5\right)}=\frac{2}{27\ln \left(5\right)}$$

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