Giải thích các bước giải:
câu 1)
\[\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-{{1}^{2}}}{x-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{(x-1)(x+1)}{x-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{x+1}{x-1} \\ & Do\left( \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{x+1}{x-1}=-\infty \ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{x+1}{x-1}=+\infty \right) \\ & \Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{x-1}=\nexists \\ \end{align}\]
câu 2)
\[\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}-x-2}{x-2}=\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)(x+1)}{x-2}=\underset{x\to 2}{\mathop{\lim }}\,(x+1)=2+1=3\]
câu 3)
\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{3}}-27}{x-3}=\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)({{x}^{2}}+3x+9)}{x-3}=\underset{x\to 3}{\mathop{\lim }}\,({{x}^{2}}+3x+9)={{3}^{2}}+3.3+9=27\]
câu 4)
\[\underset{x\to -2}{\mathop{\lim }}\,\frac{x+2}{2{{x}^{2}}-8}=\underset{x\to -2}{\mathop{\lim }}\,\frac{x+2}{2(x+2)(x-2)}=\underset{x\to -2}{\mathop{\lim }}\,\frac{1}{2(x-2)}=\frac{1}{2(-2-2)}=-\frac{1}{8}\]
câu 5)
\[\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{3}}+{{x}^{2}}-3x-3}{{{x}^{2}}-x+2}=\frac{{{(-1)}^{3}}+{{(-1)}^{2}}-3.(-1)-3}{{{(-1)}^{2}}-(-1)+2}=0\]
câu 6)
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{3+x}-2}{x-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{(x-1)(\sqrt{3+x}+2)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{\sqrt{3+x}+2}=\frac{1}{\sqrt{3+1}+2}=\frac{1}{4}\]
câu 7)
\[\begin{align} & \underset{x\to 2}{\mathop{\lim }}\,\frac{x-2}{\sqrt{3-x}-1}=\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)(\sqrt{3-x}+1)}{-(x-2)} \\ & =\underset{x\to 2}{\mathop{\lim }}\,-(\sqrt{3-x}+1)=-(\sqrt{3-2}+1)=-2 \\ \end{align}\]
câu 8)
\[\underset{x\to -3}{\mathop{\lim }}\,\frac{\sqrt{6-x}-3}{{{x}^{2}}+4x+3}=\underset{x\to -3}{\mathop{\lim }}\,\frac{-(x+3)}{(x+1)(x+3)}=\underset{x\to -3}{\mathop{\lim }}\,\frac{-1}{x+1}=\frac{-1}{-3+1}=\frac{1}{2}\]
câu 9)
\[\begin{align} & \underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{3}}-8}{2-\sqrt{2{{x}^{2}}-4}}=\underset{x\to 2}{\mathop{\lim }}\,\frac{({{x}^{3}}-8)\left( 2+\sqrt{2{{x}^{2}}-4} \right)}{-2{{x}^{2}}+8}=\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)({{x}^{2}}+2x+4)\left( 2+\sqrt{2{{x}^{2}}-4} \right)}{-2(x-2)(x+2)} \\ & =\underset{x\to 2}{\mathop{\lim }}\,\frac{({{x}^{2}}+2x+4)\left( 2+\sqrt{2{{x}^{2}}-4} \right)}{-2(x+2)}=\frac{({{2}^{2}}+2.2+4)\left( 2+\sqrt{{{2.2}^{2}}-4} \right)}{-2(2+2)}=-6 \\ \end{align}\]
câu 11)
\[\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt[3]{2-x}-1}{x-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{-(x-1)}{(x-1)\left( \sqrt[3]{{{(2-x)}^{2}}}+\sqrt[3]{2-x}+1 \right)} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{-1}{\sqrt[3]{{{(2-x)}^{2}}}+\sqrt[3]{2-x}+1}=\frac{-1}{\sqrt[3]{{{(2-1)}^{2}}}+\sqrt[3]{2-1}+1}=-\frac{1}{3} \\ \end{align}\]
câu 12)
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\sqrt[3]{1-x}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{x.\left( 1+\sqrt[3]{1-x}+\sqrt[3]{{{(1-x)}^{2}}} \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1+\sqrt[3]{1-x}+\sqrt[3]{{{(1-x)}^{2}}}}=\frac{1}{1+\sqrt[3]{1-0}+\sqrt[3]{{{(1-0)}^{2}}}}=\frac{1}{3}\]
câu 13)
\[\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt[3]{x-2}+\sqrt[3]{1-x+{{x}^{2}}}}{{{x}^{2}}-1} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{({{x}^{2}}-1)\left( \sqrt[3]{{{(x-2)}^{2}}}-\sqrt[3]{x-2}.\sqrt[3]{1-x+{{x}^{2}}}+\sqrt[3]{{{(1-x+{{x}^{2}})}^{2}}} \right)} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{\sqrt[3]{{{(x-2)}^{2}}}-\sqrt[3]{x-2}.\sqrt[3]{1-x+{{x}^{2}}}+\sqrt[3]{{{(1-x+{{x}^{2}})}^{2}}}} \\ & =\frac{1}{\sqrt[3]{{{(1-2)}^{2}}}-\sqrt[3]{1-2}.\sqrt[3]{1-1+{{1}^{2}}}+\sqrt[3]{{{(1-1+{{1}^{2}})}^{2}}}}=\frac{1}{3} \\ \end{align}\]
câu 15)
\[\begin{align} & \underset{x\to \sqrt{2}}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-2-\sqrt{2}}{{{x}^{2}}-2}=\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\frac{{{x}^{2}}-2+x-\sqrt{2}}{{{x}^{2}}-2}=\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\left( \frac{{{x}^{2}}-2}{{{x}^{2}}-2}+\frac{x-\sqrt{2}}{{{x}^{2}}-2} \right) \\ & =\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\left( 1+\frac{x-\sqrt{2}}{(x-\sqrt{2})(x+\sqrt{2})} \right)=\underset{x\to \sqrt{2}}{\mathop{\lim }}\,\left( 1+\frac{1}{x+\sqrt{2}} \right)=1+\frac{1}{\sqrt{2}+\sqrt{2}}=\frac{4+\sqrt{2}}{4} \\ \end{align}\]
