lim x->1 x+2-can x+8/can x+3 + x-3

Đáp án:

 \(\frac{-2}{3}\)

Giải thích các bước giải:

 \(lim_{x \rightarrow 1} \frac{x+2-\sqrt{x+8}}{\sqrt{x+3}+x-3}=lim_{x \rightarrow 1} \frac{[(x+2)^{2}-x-8][\sqrt{x+3}-(x-3)]}{[x+3-(x-3)^{2}][x+2+\sqrt{x+8}]}=lim_{x \rightarrow 1}\frac{(x^{2}+3x-4)[\sqrt{x+3}-(x-3)]}{(-x^{2}+7x-6)[[x+2+\sqrt{x+8}]}=lim_{x \rightarrow 1} \frac{(x-1)(x+4)[\sqrt{x+3}-(x-3)]}{(x-6)(x-1)[x+2+\sqrt{x+8}]}=lim_{x \rightarrow 1}\frac{(x+4)[\sqrt{x+3}-(x-3)]}{(x-6)[x+2+\sqrt{x+8}]}=lim_{x \rightarrow 1}\frac{(1+4)[\sqrt{1+3}-(1-3)]}{(1-6)[1+2+\sqrt{1+8}]}=\frac{-2}{3}\)