\(=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{\left(2-x\right)^2}}{\sqrt{\left(x^2+1\right)\left(2-x\right)}}=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{2-x}}{\sqrt{x^2+1}}=\frac{0}{\sqrt{5}}=0\)

\(\lim\limits_{x\rightarrow2}\dfrac{\left(3x-5\right)}{\left(x-2\right)^2}=+\infty\)

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}3x-5=3\cdot2-5=1>0\\\left(x-2\right)^2>0\\\lim\limits_{x\rightarrow2}\left(x-2\right)^2=\left(2-2\right)^2=0\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow2^-}\left(\dfrac{1}{x-2}-\dfrac{1}{x^2-4}\right)\)

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

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

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2^-}\dfrac{x+1}{x+2}=\dfrac{2+1}{2+2}=\dfrac{3}{4}>0\\x-2< 0\end{matrix}\right.\)

\(=\frac{3\sqrt{3}}{0^+}=+\infty\)

Câu dưới là 1 giới hạn hoàn toàn bình thường (không phải dạng vô định), bạn cứ thay số vào là được thôi

\(\lim\limits_{x\rightarrow0}\left(1-x\right)tan\frac{\pi x}{2}=\left(1-0\right).tan0=1\)

giai cau duoi thoi nha

Giới hạn đã cho hữu hạn nên \(x^2+2ax-b=0\) có nghiệm \(x=2\)

\(\Rightarrow4+4a-b=0\Rightarrow b=4a+4\)

\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{x^2+2ax-4a-4}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2a+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x+2a+2}{x+2}=\dfrac{2a+4}{4}=4\)

\(\Rightarrow a=6\Rightarrow b=28\)