\(\lim\limits_{x\rightarrow2}\dfrac{\left(3x-5\right)}{\left(x-2\right)^2}=+\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}3x-5=3\cdot2-5=1>0\\\left(x-2\right)^2>0\\\lim\limits_{x\rightarrow2}\left(x-2\right)^2=\left(2-2\right)^2=0\end{matrix}\right.\)
\(lim_{x\rightarrow2^-}\left(\dfrac{1}{x-2}-\dfrac{1}{x^2-4}\right)\)
\(lim_{x\rightarrow2^-}\frac{x^2-4}{\sqrt{\left(x^2+1\right)\left(2-x\right)}}\)
\(lim_{x\rightarrow2^-}\frac{\left|x-2\right|}{x-2}\)
Tìm giới hạn :
\(lim_{x\rightarrow2^+}\dfrac{x^2-3x+3}{x-2}\)
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
\(lim_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\left(\sqrt{\dfrac{3x}{x^2-1}}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+3x\right)^3-\left(1-4x\right)^4}{x}\)
\(\lim\limits_{x\rightarrow2}\dfrac{2x^2-5x+2}{x^3-3x-2}\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^4-3x+2}{x^3+2x-3}\)
1) \(\overset{lim}{x\rightarrow1}\)\(\dfrac{x^3-3x+2}{x^4-4x+3}\)\(\)
2)\(\overset{lim}{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}\)
3) \(\overset{lim}{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}\)
4)\(\overset{lim}{x\rightarrow0^-}\dfrac{\left(1+x\right)\left(1+4x\right)-1}{x^2}\)
5) \(\overset{lim}{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}\)
\(lim_{x\rightarrow0}\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)\)
