Помогите с решением предела [tex] \lim_{x \to 3}( \frac{6}{ x^{2} -9}- \frac{1}{x-3})[/tex]. Created by bangbang635. algebra-ru

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July 2022 2 8 Report

Помогите с решением предела
[tex] \lim_{x \to 3}( \frac{6}{ x^{2} -9}- \frac{1}{x-3})[/tex]

Змей24 $\lim_{x \to 3} \bigg ( \dfrac{6}{x^2 - 9} - \dfrac{1}{x - 3} \bigg ) = \lim_{x \to 3} \bigg ( \dfrac{6}{(x - 3)(x + 3)} - \dfrac{1}{x - 3} \bigg) = \\ \\ 
\lim_{x \to 3} \dfrac{6 - x - 3 }{(x - 3)(x + 3)} = \lim_{x \to 3} \dfrac{-(x - 3)}{(x - 3)(x + 3)} = \\ \\ 
= \lim_{x \to 3} \dfrac{-1}{x + 3 } = -\dfrac{1}{3 + 3} = -\dfrac{1}{6}$

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Verified answer

Lim[x->3](6/(x^2 - 9) - 1/(x-3)) = lim[x->3](6/(x^2 - 9) - (x+3)/(x^2 - 9)) = lim[x->3]((6 - x - 3)/(x^2 - 9)) = lim[x->3]((3 - x)/((x-3)*(x+3)))) = lim[x->3](-1/(x+3)) = -1/6

Ответ: -1/6

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