Решите пожалуйста лимит [tex]\lim_{x \to7} \frac{2-\sqrt{x-3} }{x^{2} + x -56}[/tex]. Created by lepme. algebra-ru

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$\lim\limits _{x \to 7}\frac{2-\sqrt{x-3}}{x^2+x-56}=\lim\limits _{x \to 7}\frac{(2-\sqrt{x-3})(2+\sqrt{x-3})}{(x-7)(x+8)(2+\sqrt{x-3})}=\\\\=\lim\limits _{x \to 7}\frac{4-(x-3)}{(x-7)(x+8)(2+\sqrt{x-3})}=\lim\limits _{x \to 7}\frac{-(x-7)}{(x-7)(x+8)(2+\sqrt{x-3})}=\\\\=\lim\limits _{x \to 7}\frac{-1}{(x+8)(2+\sqrt{x-3})}=\frac{-1}{15\cdot 4}=-\frac{1}{60}$

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Verified answer

lim(→7) ((2-√(x-3))/(x²+x-56))

Неопределённость 0/0.

Берём производную одновременно от с числителя и знаменателя:

(2-√(x-3))`/(x²+x-56)`=(-1/(2*√(x-3)))/(2x+1)=-1/(2*√(x-3)*(2x+1)). ⇒

lim(x→7) (-1/(2*√(x-3)*(2x+1)))=-1/(2*√4*(2*7+1))=-1/(2*2*15)=-1/60.

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