\(\text { Tính giới hạn } A=\lim\limits _{x \rightarrow 1} \frac{x^{2}+3 x+2-2 \sqrt{6 x^{2}+3 x}}{x^{2}-2 x+2-\cos (x-1)} \text { . }\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } A=\lim \limits _{x \rightarrow 1} \frac{(x+1)(x+2-\sqrt{6 x+3})+\sqrt{6 x+3}(x+1-2 \sqrt{x})}{(x-1)^{2}+2 \sin ^{2} \frac{x-1}{2}} \\ &=\lim \limits _{x \rightarrow 1} \frac{\frac{(x+1)\left[(x+2)^{2}-(6 x+3)\right]}{x+2+\sqrt{6 x+3}}+\sqrt{6 x+3} \cdot \frac{(x+1)^{2}-4 x}{x+1+2 \sqrt{x}}}{(x-1)^{2}+2 \sin ^{2} \frac{x-1}{2}}=\lim \limits _{x \rightarrow 1} \frac{\frac{(x-1)^{2}(x+1)}{x+2+\sqrt{6 x+3}}+\frac{(x-1)^{2} \sqrt{6 x+3}}{x+1+2 \sqrt{x}}}{(x-1)^{2}+2 \sin ^{2} \frac{x-1}{2}} \\ &=\lim \limits _{x \rightarrow 1} \frac{\frac{x+1}{x+2+\sqrt{6 x+3}}+\frac{\sqrt{6 x+3}}{x+1+2 \sqrt{x}}}{1+2 \frac{\sin ^{2} \frac{x-1}{2}}{(x-1)^{2}}}=\lim \limits _{x \rightarrow 1} \frac{\frac{x+1}{x+2+\sqrt{6 x+3}}+\frac{\sqrt{6 x+3}}{x+1+2 \sqrt{x}}}{1+\frac{1}{2}\left(\frac{\sin \frac{x-1}{2}}{\frac{x-1}{2}}\right)^{2}}=\frac{\frac{2}{6}+\frac{3}{4}}{1+\frac{1}{2}}=\frac{13}{18} . \end{aligned} \)