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Question Number 98213 by me2love2math last updated on 12/Jun/20

Commented by bemath last updated on 12/Jun/20

$$\mathrm{let}\frac{1}{\mathrm{x}}=\mathrm{t},\mathrm{t}\to 0$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{\ln (1+3t)}{t}=\underset{\mathrm{t}\to 0}{\lim }\frac{\frac{3}{1+3\mathrm{t}}}{1}=\frac{3}{1}=3$$

Answered by Rio Michael last updated on 12/Jun/20

$$\underset{x\to \mathrm{\infty }}{\lim }\left[x\ln (1+\frac{3}{x})\right]=\underset{x\to \mathrm{\infty }}{\lim }\left[x\ln \left(\frac{x+3}{x}\right)\right]$$$$\mathrm{By}{\mathrm{l}}^{\prime }\mathrm{hopitals}\mathrm{rule}\underset{x\to \mathrm{\infty }}{\lim }\frac{2x-3}{3+x}+\underset{x\to }{\lim }x\ln (1+\frac{3}{x})=3+0=3$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\left[x\ln (1+\frac{3}{x})\right]=3$$$$$$

Answered by MJS last updated on 12/Jun/20

$$\underset{x\to +\mathrm{\infty }}{\lim }\left(x\ln (1+\frac{3}{x})\right)\stackrel{[x=\frac{1}{t}]}{=}\underset{t\to 0^{+}}{\lim }\frac{\ln (3t+1)}{t}=$$$$=\underset{t\to 0^{+}}{\lim }\frac{\frac{d}{dt}\left[\ln (3t+1)\right]}{\frac{d}{dt}\left[t\right]}=\underset{t\to 0^{+}}{\lim }\frac{3}{3t+1}=3$$

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