If-lim-x-0-ln-a-x-ln-a-x-k-lim-x-e-ln-x-1-x-e-1-then-k-

Question Number 128411 by bramlexs22 last updated on 07/Jan/21

If \lim_{x \to 0} (\frac{\ln (a + x) - \ln a}{x}) + k \lim_{x \to e} (\frac{\ln x - 1}{x - e}) = 1

then k =

Answered by Dwaipayan Shikari last updated on 07/Jan/21

\lim_{x \to 0} \frac{l o g (1 + \frac{x}{a})}{x} + \frac{k}{e} \lim_{x \to e} \frac{l o g (\frac{x}{e})}{\frac{x}{e} - 1}

= \frac{\frac{x}{a}}{x} + \frac{k}{e} . \frac{l o g (1 + \frac{x}{e} - 1)}{\frac{x}{e} - 1} \Rightarrow \frac{1}{a} + \frac{k}{e} = 1 \Rightarrow k = e (1 - \frac{1}{a}) \lim_{x \to 0} l o g (1 + x) = x

Answered by mr W last updated on 07/Jan/21

\frac{1}{a + 0} + k \frac{1}{e} = 1

\Rightarrow k = (1 - \frac{1}{a}) e

Answered by liberty last updated on 07/Jan/21

L^{'} Hopital

\lim_{x \to 0} \frac{\frac{1}{a + x}}{1} + k \lim_{x \to e} \frac{\frac{1}{x}}{1} = 1

\frac{1}{a} + \frac{k}{e} = 1 \Rightarrow k = \frac{a e - e}{a} = \frac{e}{a} (a - 1)

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