Question

Lim x> 1 x^2-корень из x/корень и x -1

Answer 1

$\displaystyle \lim_{x \to \1} \frac{x^2- \sqrt{x}}{ \sqrt{x-1}}= \lim_{x \to \1} \sqrt{ \frac{(x^2- \sqrt{x})^2}{x-1}}= \lim_{x \to \1} \sqrt{ \frac{(x^2- \sqrt{x})^2(x^2+ \sqrt{x} )}{(x-1)(x^2+ \sqrt{x} )}}= \\\\= \lim_{x \to \1} \sqrt{ \frac{(x^4-2x^2 \sqrt{x} +x)(x^2+ \sqrt{x} )}{(x-1)(x+ \sqrt{x} )}}=\\\\= \lim_{x \to \1} \sqrt{ \frac{x^6-2x^4 \sqrt{x} +x^2+x^4 \sqrt{x} -2x^3+x \sqrt{x} }{(x-1)(x^2+ \sqrt{x} )}}=\\\\$
$\displaystyle = \lim_{x \to \1} \sqrt{ \frac{x^6-x^4 \sqrt{x} -x^3+x \sqrt{x} }{(x-1)(x^2+ \sqrt{x} )}}= \lim_{x \to \1} \sqrt{ \frac{x^4(x^2- \sqrt{x})-x(x^2- \sqrt{x} )}{(x-1)(x^2+ \sqrt{x} )}}=\\\\= \lim_{x \to \1} \sqrt{ \frac{(x^4-x)(x^2- \sqrt{x} )}{(x-1)(x^2+ \sqrt{x} )}}= \lim_{x \to \1} \sqrt{ \frac{x(x^3-1)(x^2- \sqrt{x} )}{(x-1)(x^2+ \sqrt{x} )}}=\\\\= \lim_{x \to \1} \sqrt{ \frac{x(x-1)(x^2+x+1)(x^2- \sqrt{x} )}{(x-1)(x^2+ \sqrt{x} )}} =$
$\displaystyle = \lim_{x \to \1} \sqrt{ \frac{x(x^2+x+1)(x^2- \sqrt{x} )}{(x^2+ \sqrt{x} )}}= \sqrt{ \frac{1*3*0}{2}}=0$

Answer 2

$\displaystyle \lim_{x \to 1} \frac{x^2-\sqrt{x}}{\sqrt{x-1}}= \lim_{x \to 1} \frac{(x^2-\sqrt{x})(x^2+\sqrt{x})}{\sqrt{x-1}(x^2+\sqrt{x})}=\lim_{x \to 1} \frac{x^4-x}{\sqrt{x-1}(x^2+\sqrt{x})}=\\\\\\=\lim_{x \to 1} \frac{x(x^3-1)}{\sqrt{x-1}(x^2+\sqrt{x})}=\lim_{x \to 1} \frac{x(x-1)(x^2+x+1)}{\sqrt{x-1}(x^2+\sqrt{x})}=\\\\\\=\lim_{x \to 1} \frac{x\sqrt{x-1}(x^2+x+1)}{(x^2+\sqrt{x})}=\frac{1\sqrt{1-1}(1^2+1+1)}{1^2+\sqrt{1}}=\frac{0}2=0$