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$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{x}}}{\left(2^{\frac{1}{x}}-1\right)x}\right)$

I've tried a couple of things but I can't get to the right result.. according to online calculators the limit equals $\frac{1}{ln\left(2\right)}$ but I keep getting 0. Is there a a trick to solve it?

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    $\begingroup$ Take the limits for the numerator and the denominator separately. $\endgroup$ Apr 12 at 11:32
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    $\begingroup$ expand numerator and denumerator with $2^x$ $\endgroup$
    – lmaosome
    Apr 12 at 11:38
  • $\begingroup$ Thanks for the help guys! :) $\endgroup$ Apr 12 at 11:45
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$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{x}}}{\left(2^{\frac{1}{x}}-1\right)x}\right)$

taking $1/x$=$t$ as $x$ tends to $\infty$, $t$ tends to $0$

then apply L-hopital rule or use $lim_{t\to0}\frac{2^t-1}{t}$ = $ln2$

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  • $\begingroup$ Thank you so much I completely forgot about substitution method! :) $\endgroup$ Apr 12 at 11:45

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