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Question Number 184960 by mathlove last updated on 14/Jan/23

Answered by cortano1 last updated on 15/Jan/23

$$L=-\frac{9}{4}(\ln 4-1)$$

Commented by mathlove last updated on 15/Jan/23

$$solution???$$

Answered by cortano1 last updated on 15/Jan/23

$$\underset{x\to 1}{\lim }\frac{\ln (x+1)-\ln 2}{\sqrt[3]{7+x}-\sqrt{1+3x}}\times \underset{x\to 1}{\lim }\frac{3(4^{x-1}-x)}{\sin (x-1)}$$$$L_1=\underset{t\to 0}{\lim }\frac{\ln (t+2)-\ln 2}{\sqrt[3]{t+8}-\sqrt{4+3t}}$$$$L_1=\underset{t\to 0}{\lim }\frac{\frac{1}{t+2}}{\frac{1}{3\sqrt[3]{{(t+8)}^2}}-\frac{3}{2\sqrt{4+3t}}}$$$$L_1=\frac{\frac{1}{2}}{\frac{1}{12}-\frac{3}{4}}=\frac{6}{1-9}=-\frac{3}{4}$$$$L_2=\underset{x\to 1}{\lim }\frac{3(4^{x-1}-x)}{\sin (x-1)}=3.\underset{x\to 1}{\lim }\frac{4^{x-1}-1-(x-1)}{x-1}$$$$L_2=3(\ln 4-1)$$$$\therefore L=L_1\times L_2=-\frac{9}{4}(\ln 4-1)$$

Commented by mathlove last updated on 15/Jan/23

$$thanksdear$$

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