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Question Number 99935 by student work last updated on 24/Jun/20

lim_(x→0) (1/x^(ln(e^x −1)) )=?  help me

$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)} }=? \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$

Commented by student work last updated on 24/Jun/20

help me sisrs

$$\mathrm{help}\:\mathrm{me}\:\mathrm{sisrs} \\ $$

Commented by student work last updated on 24/Jun/20

$$ \\ $$

Commented by student work last updated on 24/Jun/20

who can help me?

$$\mathrm{who}\:\mathrm{can}\:\mathrm{help}\:\mathrm{me}? \\ $$

Commented by student work last updated on 24/Jun/20

l need your help

$$\mathrm{l}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help} \\ $$

Answered by bobhans last updated on 24/Jun/20

lim_(x→0)  x^(−ln(e^x −1))  = lim_(x→0)  ((1/x))^(ln(e^x −1))   = e^(lim_(x→0)  ln(e^x −1). ln((1/x)))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{x}^{−\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)} \:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)} \\ $$$$=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right).\:\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \\ $$$$ \\ $$

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