Question-217733

Question Number 217733 by Samuel12 last updated on 19/Mar/25

Answered by vnm last updated on 19/Mar/25

x^{\frac{1}{\ln (e^{x} - 1)}} = e^{\frac{\ln x}{\ln (e^{x} - 1)}} = e^{\frac{\ln x}{\ln x + \ln \frac{e^{x} - 1}{x}}} =

e^{\frac{1}{1 + \frac{1}{\ln x} \ln \frac{e^{x} - 1}{x}}} \underset{x \to 0 +}{\to} e^{\frac{1}{1 + \frac{1}{- \infty} \cdot \ln 1}} = e

Answered by vnm last updated on 19/Mar/25

{(\ln (1 + e^{- x}))}^{\frac{1}{x}} = {(\frac{1}{e^{x}} \ln {(1 + e^{- x})}^{e^{x}})}^{\frac{1}{x}} =

e^{- 1} {(\ln {(1 + \frac{1}{e^{x}})}^{e^{x}})}^{\frac{1}{x}} \underset{x \to + \infty}{\to} e^{- 1} {(\ln e)}^{0} = e^{- 1}

Answered by vnm last updated on 20/Mar/25

\frac{e^{x} - e^{2}}{x^{2} + x - 6} \underset{x = 2 + t}{=} \frac{e^{2} (e^{t} - 1)}{{(2 + t)}^{2} + 2 + t - 6} = \frac{e^{2} (e^{t} - 1)}{t (t + 5)} =

\frac{e^{2}}{t + 5} \cdot \frac{e^{t} - 1}{t} \underset{t \to 0}{\to} \frac{e^{2}}{5} \cdot 1 = \frac{e^{2}}{5}

\frac{\tan x - \sin x}{\sin x (\cos 2 x - \cos x)} = \frac{\frac{1}{\cos x} - 1}{\cos 2 x - \cos x} =

\frac{1 - \cos x}{\cos x ({2 \cos}^{2} x - 1 - \cos x)} =

\frac{1}{\cos x} \cdot \frac{1 - \cos x}{(2 \cos x + 1) (\cos x - 1)} =

\frac{1}{\cos x} \cdot \frac{- 1}{2 \cos x + 1} \underset{x \to 0}{\to} - \frac{1}{3}

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