I can't find the way to solve this question and i always get 0/0. The question is: $\lim_{x\rightarrow \infty } x\left [ 2^{\frac{1}{x}}-1 \right ]$
From Mathematica, i get -infinity. But how can it done using manual method...? Thanks
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Setting $x=\dfrac1h$
we have $$\lim_{h\to0}\frac{2^h-1}h$$
$$=\lim_{h\to0}\frac{(e^{\ln 2})^h-1}h$$
$$=\ln2\cdot\lim_{h\to0}\frac{(e^{h\ln 2})-1}{h\ln2}=?$$
lab bhattacharjee
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Oh... So we can set x become 1/h? – user3658777 Nov 19 '14 at 10:24
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I think this also get infinity... – user3658777 Nov 19 '14 at 10:26
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@user3658777, Do you know : http://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1 – lab bhattacharjee Nov 19 '14 at 10:27
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it gets infinity, but from wolfram alpha and mathematica, it gets -infinity – user3658777 Nov 19 '14 at 10:31
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@user3658777, It should be $\ln2\cdot1$ – lab bhattacharjee Nov 19 '14 at 11:06