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Question Number 167663 by mathlove last updated on 22/Mar/22

$$\underset{x\to 2}{\lim }\left[log(\frac{1}{x}+\frac{1}{2x}+\frac{1}{4x}.......)\right]=?$$

Commented by mathlove last updated on 22/Mar/22

$$????$$

Commented by alephzero last updated on 22/Mar/22

$$didYoumean$$$$\frac{1}{x}+\frac{1}{2x}+\frac{1}{4x}+...=\frac{1}{x}+\frac{1}{2x}+...+\frac{1}{2nx}+...$$$$or$$$$\frac{1}{x}+\frac{1}{2x}+\frac{1}{4x}+...=\frac{1}{x}+\frac{1}{2x}+...+\frac{1}{2^{n}x}+...?$$

Answered by alephzero last updated on 22/Mar/22

$$\frac{1}{2^{0}x}+\frac{1}{2^{1}x}+\frac{1}{2^{2}x}+...=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{n}x}$$$$S_n=\frac{a_1(1-r^n)}{1-r}$$$$a_1=\frac{1}{x}\wedge r=\frac{1}{2}$$$$\Rightarrow S=S_{\mathrm{\infty }}=\frac{\left(\frac{1}{x}\right)}{\left(\frac{1}{2}\right)}=\frac{2}{x}$$$$\Rightarrow \underset{x\to 2}{\lim ln}(\frac{1}{x}+\frac{1}{2x}+...)=$$$$=\underset{x\to 2}{\lim ln}\left(\frac{2}{x}\right)=0$$

Commented by mathlove last updated on 22/Mar/22

$$thankssir$$

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