lim-x-1-1-ln-x-1-x-1-

Question Number 131962 by EDWIN88 last updated on 10/Feb/21

\lim_{x \to 1^{+}} (\frac{1}{\ln x} - \frac{1}{x - 1}) = ?

Answered by liberty last updated on 10/Feb/21

let u = \ln x \Rightarrow x = e^{u}

\lim_{u \to 0} (\frac{1}{u} - \frac{1}{e^{u} - 1})

= \lim_{u \to 0} (\frac{e^{u} - 1 - u}{u (e^{u} - 1)})

= \lim_{u \to 0} (\frac{e^{u} - 1}{e^{u} - 1 + {ue}^{u}})

= \lim_{u \to 0} (\frac{e^{u}}{e^{u} + e^{u} + {ue}^{u}}) = \frac{1}{2}

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