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Question Number 98539 by Rio Michael last updated on 14/Jun/20

Given the function

f (x) = \frac{\ln x}{x - 1}

(a) State the domain D_{f} of f .

(b) Find \lim_{x \to \infty} \frac{\ln x}{x - 1} . State its asymptotes .

(c) Draw up a variation table for the curve y = f (x) .

Answered by Aziztisffola last updated on 14/Jun/20

a) D_{f} =] 0; 1 [\cup] 1; + \infty [

b) \underset{x \to \infty}{\lim f} (x) = 0

Commented by Rio Michael last updated on 14/Jun/20

great sir, is x = 0 an asymptote or

x = 1 ? ? which

Commented by Aziztisffola last updated on 14/Jun/20

yes x = 0 and x = 1

Answered by mathmax by abdo last updated on 14/Jun/20

1) f is defined on] 0, 1 [\cup] 1, + \infty [

2) \lim_{x \to + \infty} \frac{lnx}{x - 1} = \lim_{x \to + \infty} \frac{lnx}{x} \times \frac{1}{1 - \frac{1}{x}} = \lim_{x \to + \infty} \frac{lnx}{x} = 0 so y = 0 is

assymptote alway x = 1 is assymtote .

3) f^{^{'}} (x) = \frac{\frac{1}{x} (x - 1) - lnx}{{(x - 1)}^{2}} = \frac{(x - 1) - xlnx}{x {(x - 1)}^{2}} = \frac{φ (x)}{x {(x - 1)}^{2}}

φ (x) = x - 1 - xlnx \Rightarrow φ^{^{'}} (x) = 1 - (lnx + 1) = - lnx

φ^{^{'}} > 0 \Leftrightarrow - lnx > 0 \Leftrightarrow lnx < 0 \Leftrightarrow 0 < x < 1

x 0 1 + \infty

φ^{^{'}} ∣∣ + -

φ ∣∣ - \infty decr 0 decr - \infty \Rightarrow φ (x) ⩽ 0 \Rightarrow f^{^{'}} < 0 \Rightarrow f is strictly ideccreazing

vareiation of f

x 0 1 + \infty \lim_{x \to 1} f (x) = 1

f^{^{'}} (x) ∣∣ - ∣∣ -

f (x) ∣∣ + \infty dec 1 decr 0

Commented by Rio Michael last updated on 14/Jun/20

Brilliant work you two .

thanks

Commented by abdomathmax last updated on 15/Jun/20

you are welcome