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Question Number 192969 by mustafazaheen last updated on 01/Jun/23

Commented by mustafazaheen last updated on 01/Jun/23

$$\mathrm{how}\mathrm{is}\mathrm{solution}?$$

Answered by MM42 last updated on 01/Jun/23

$${lim}_{x\to 1^{+}}{\left(\frac{3}{x-1}\right)}^{ln\frac{4}{3}}={\left(\frac{3}{0^{+}}\right)}^{+}={(+\mathrm{\infty })}^{+}=+\mathrm{\infty }$$

Answered by aba last updated on 01/Jun/23

$${(\frac{\mathrm{x}}{3}-\frac{1}{3})}^{\ln \left(\frac{3}{4}\right)}={\mathrm{e}}^{\ln \left(\frac{3}{4}\right)\ln \left(\frac{1}{3}\right)+\ln \left(\frac{3}{4}\right)\ln (\mathrm{x}-1)}$$$$\underset{x\to 1^{+}}{\lim }{(\frac{\mathrm{x}}{3}-\frac{1}{3})}^{\ln \left(\frac{3}{4}\right)}={\mathrm{e}}^{+\mathrm{\infty }}=+\mathrm{\infty }$$

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