\(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}\)

ta có : \(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{ }=1\) và \(\dfrac{lim}{x\rightarrow2}\dfrac{x-2}{ }=0\)

\(\Rightarrow\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}=\infty\)

và \(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}=+\infty\) khi \(x>2\)

bỏ giùm mk cái trường hợp \(x< 2\) đi nha

\(\lim\limits_{x\rightarrow2}\dfrac{2x^3+5x^2-7x+2}{x^2-3x+2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2x^3-4x^2+9x^2-18x+11x-22+24}{\left(x-2\right)\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(2x^2+9x+11\right)+24}{\left(x-2\right)\left(x+1\right)}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\left(x-2\right)\left(2x^2+9x+11\right)+24=24>0\\\lim\limits_{x\rightarrow2}\left(x-2\right)=2-2=0\\\lim\limits_{x\rightarrow2}x+1=2+1=3>0\end{matrix}\right.\)

Sai.

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)

Bạn tự hiểu là giới hạn khi x tới 2:

\(=\frac{x\left(x-2\right)\left[2\sqrt{x+2}+3x-2\right]}{4\left(x+2\right)-\left(3x-2\right)^2}=\frac{x\left(x-2\right)\left[2\sqrt{x+2}+3x-2\right]}{-9x^2+16x+4}=\frac{x\left(x-2\right)\left[2\sqrt{x+2}+3x-2\right]}{\left(x-2\right)\left(-9x-2\right)}\)

\(=\frac{x\left[2\sqrt{x+2}+3x-2\right]}{-9x-x}=\frac{2\left[2\sqrt{4}+6-2\right]}{-18-2}=...\)

Xét giới hạn \(L=\lim\limits_{x\rightarrow2}\frac{x^2-5x+6}{x^3-x^2-x-2}\)

                         \(=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow2}\frac{x-3}{x^2+x+1}=-\frac{1}{7}\)

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)

\(L=\lim\limits_{x\rightarrow2}\frac{x-\sqrt{3x-2}}{x^2-4}\)

   \(=\lim\limits_{x\rightarrow2}\frac{x^2-3x+2}{\left(x-4\right)\left(x+\sqrt{3x-2}\right)}=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)\left(x+\sqrt{3x-2}\right)}\)

   \(=\lim\limits_{x\rightarrow2}\frac{x-1}{\left(x+2\right)\left(x+\sqrt{3x-2}\right)}=\frac{1}{16}\)

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