lim-x-2-log-1-x-1-2x-1-4x-

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

\lim_{x \to 2} [l o g (\frac{1}{x} + \frac{1}{2 x} + \frac{1}{4 x} \dots \dots .)] = ?

Commented by mathlove last updated on 22/Mar/22

? ? ? ?

Commented by alephzero last updated on 22/Mar/22

d i d Y o u m e a n

\frac{1}{x} + \frac{1}{2 x} + \frac{1}{4 x} + \dots = \frac{1}{x} + \frac{1}{2 x} + \dots + \frac{1}{2 n x} + \dots

o r

\frac{1}{x} + \frac{1}{2 x} + \frac{1}{4 x} + \dots = \frac{1}{x} + \frac{1}{2 x} + \dots + \frac{1}{2^{n} x} + \dots ?

Answered by alephzero last updated on 22/Mar/22

\frac{1}{2^{0} x} + \frac{1}{2^{1} x} + \frac{1}{2^{2} x} + \dots = \underset{n = 0}{\sum^{\infty}} \frac{1}{2^{n} x}

S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}

a_{1} = \frac{1}{x} \land r = \frac{1}{2}

\Rightarrow S = S_{\infty} = \frac{(\frac{1}{x})}{(\frac{1}{2})} = \frac{2}{x}

\Rightarrow \underset{x \to 2}{\lim l n} (\frac{1}{x} + \frac{1}{2 x} + \dots) =

= \underset{x \to 2}{\lim l n} (\frac{2}{x}) = 0

Commented by mathlove last updated on 22/Mar/22

t h a n k s s i r