Tìm lim

Question

$\mathop {\lim }_{x \to 1} $$\frac{x^{2}-4x+3}{\sqrt{4x+5}-3}$

Dân Nguyễn · Answer 1 · 12/26/2014 10:56:42 PM

$\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x-3)}{\sqrt{4x+5}-3}=\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x-3)(\sqrt{4x+5}+3)}{4(x-1)}$

$=\mathop {\lim }\limits_{x \to 1}\frac{(x-3)(\sqrt{4x+5}+3)}{4}=\mathop {\lim }\limits_{x \to 1}\frac{-2.6}{4}=-3$

Trả lời 26-12-14 10:56 PM

Dân Nguyễn
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score 1 · Answer 2

$\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x-3)}{\sqrt{4x+5}-3}=\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x-3)(\sqrt{4x+5}+3)}{4(x-1)}$

$=\mathop {\lim }\limits_{x \to 1}\frac{(x-3)(\sqrt{4x+5}+3)}{4}=\mathop {\lim }\limits_{x \to 1}\frac{-2.6}{4}=-3$

2 Đáp án