题目
lim x → 2 x − 2 + x − 2 x 2 − 4 \lim_{x\to 2}\frac{\sqrt{x}-\sqrt{2}+\sqrt{x-2}}{\sqrt{x^2-4}} x→2limx2−4x−2+x−2
解答
lim x → 2 x − 2 + x − 2 x 2 − 4 = lim x → 2 x − 2 + x − 2 x − 2 x + 2 = 1 2 lim x → 2 x − 2 + x − 2 x − 2 = 1 2 ( lim x → 2 x − 2 x − 2 + 1 ) = 1 2 lim x → 2 ( x − 2 ) ( x + 2 ) x − 2 ( x + 2 ) + 1 2 = 1 2 lim x → 2 ( x − 2 ) 2 x − 2 ( x + 2 ) + 1 2 = 1 2 lim x → 2 x − 2 x + 2 + 1 2 = 1 2 \begin{aligned} \lim_{x\to 2}\frac{\sqrt{x}-\sqrt{2}+\sqrt{x-2}}{\sqrt{x^2-4}}&=\lim_{x\to 2}\frac{\sqrt{x}-\sqrt{2}+\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}}\\[8pt] &=\frac{1}{2}\lim_{x\to 2}\frac{\sqrt{x}-\sqrt{2}+\sqrt{x-2}}{\sqrt{x-2}}\\[8pt] &=\frac{1}{2}(\lim_{x\to 2}\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}+1)\\[8pt] &=\frac{1}{2}\lim_{x\to 2}\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{\sqrt{x-2}(\sqrt{x}+\sqrt{2})}+\frac{1}{2}\\[8pt] &=\frac{1}{2}\lim_{x\to 2}\frac{(\sqrt{x-2})^2}{\sqrt{x-2}(\sqrt{x}+\sqrt{2})}+\frac{1}{2}\\[8pt] &=\frac{1}{2}\lim_{x\to 2}\frac{\sqrt{x-2}}{\sqrt{x}+\sqrt{2}}+\frac{1}{2}\\[8pt] &=\frac{1}{2}\\[8pt] \end{aligned} x→2limx2−4x−2+x−2=x→2limx−2x+2x−2+x−2=21x→2limx−2x−2+x−2=21(x→2limx−2x−2+1)=21x→2limx−2(x+2)(x−2)(x+2)+21=21x→2limx−2(x+2)(x−2)2+21=21x→2limx+2x−2+21=21
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