Предел:
$lim_{x to 0} frac{x}{ sqrt[4]{1+2x} -1} =$

$lim_{x to 0} frac{ frac{dx}{dx} }{ frac{d}{dx}( sqrt[4]{2x+1}-1) } =$ $lim_{x to 0} frac{1}{ frac{1}{2(2x+1)^{ }} } = lim_{x to 0} 2(2x+1)^{3/4}=2(2*0+1)^{3/4}=2$
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$lim_{t to 1} frac{ frac{t^{16}-1}{2} }{t^4-1} = lim_{t to 1} frac{{t^{16}-1} }{2(t^4-1)}=lim_{t to 1} frac{{(t^{8}-1)(t^{8}+1)} }{2(t^4-1)}$ $lim_{t to 1} frac{{(t^{4}-1)(t^{4}+1)(t^{8}+1)} }{2(t^4-1)}=lim_{t to 1} frac{{((t^{4}+1)(t^{8}+1)} }{2}= lim_{t to 1} frac{2*2}{2}=2$
-------------------------Объяснение------------------
$sqrt[4]{1+2x} =t^4$ ;
$t^4= sqrt[4]{1+2x}$
$t^{16}=1+2x$
$x= frac{t^{16}-1}{2}$
Почему поменялся предел:
$frac{t^{16}-1}{2} to 0$
$t^{16}-1 to 0$
$t^{16} to 1$
$t to 1$