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Lời giải:
\(\frac{x-2}{\sqrt{5x-1}+\sqrt{x+2}-5}=\frac{x-2}{(\sqrt{5x-1}-3)+(\sqrt{x+2}-2)}=\frac{x-2}{\frac{5(x-2)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}}\)
Do đó:
\(\lim_{x\to 2}\frac{x-2}{\sqrt{5x-1}+\sqrt{x+2}-5}=\lim_{x\to 2}\frac{x-2}{\frac{5(x-2)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}}=\lim_{x\to 2}\frac{1}{\frac{5}{\sqrt{5x-1}+3}+\frac{1}{\sqrt{x+2}+2}}=\frac{12}{13}\)
\(L=\lim\limits_{x\rightarrow2}\frac{x-\sqrt{3x-2}}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\frac{x^2-3x+2}{\left(x-4\right)\left(x+\sqrt{3x-2}\right)}=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)\left(x+\sqrt{3x-2}\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{x-1}{\left(x+2\right)\left(x+\sqrt{3x-2}\right)}=\frac{1}{16}\)
a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)
b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)
\(a=\frac{0-1}{0-1}=1\)
\(b=\lim\limits_{x\rightarrow0}\frac{\frac{x^2}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}}{x^2}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}=\frac{1}{3}\)
\(c=\lim\limits_{x\rightarrow2}\frac{\sqrt{x+2}-2+\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{\frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{x+7}+3}}{x-2}=\lim\limits_{x\rightarrow2}\left(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{x+7}+3}\right)\)
\(=\frac{1}{\sqrt{4}+2}+\frac{1}{\sqrt{9}+3}=\frac{5}{12}\)
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:
\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)
\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)
\(c=0.sin\frac{1}{2}=0\)
Bạn tự hiểu là lim:
\(=\frac{\left(x^2-x-2\right)}{\left(4x-8\right)}.\frac{\left(\sqrt{4x+1}+3\right)}{\left(x+\sqrt{x+2}\right)}=\frac{\left(x+1\right)\left(x-2\right)}{4\left(x-2\right)}.\frac{\left(\sqrt{4x+1}+3\right)}{\left(x+\sqrt{x+2}\right)}=\frac{\left(x+1\right)}{4}.\frac{\left(\sqrt{4x+1}+3\right)}{\left(x+\sqrt{x+2}\right)}\)
\(=\frac{3\left(\sqrt{9}+3\right)}{4\left(2+\sqrt{4}\right)}=...\)
Xét giới hạn \(L=\lim\limits_{x\rightarrow2}\frac{x^2-5x+6}{x^3-x^2-x-2}\)
\(=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow2}\frac{x-3}{x^2+x+1}=-\frac{1}{7}\)