đáp án:$\lim _{x\to \infty \:}\left(\frac{2x^3+2}{\sqrt{2x-3}+x-\frac{1}{2}}\right)=\lim _{x\to \infty \:}\left(\frac{2x^2+\frac{2}{x}}{\frac{\sqrt{2x-3}}{x}+1-\frac{1}{2x}}\right)=\frac{\lim _{x\to \infty \:}\left(2x^2+\frac{2}{x}\right)}{\lim _{x\to \infty \:}\left(\frac{\sqrt{2x-3}}{x}+1-\frac{1}{2x}\right)}=\frac{\infty \:}{1}=\infty \: giải thích:(\lim _{x\to \infty \:}\left(2x^2+\frac{2}{x}\right)=\lim _{x\to \infty \:}\left(2x^2\right)+\lim _{x\to \infty \:}\left(\frac{2}{x}\right)=\infty \:+0; \lim _{x\to \infty \:}\left(\frac{\sqrt{2x-3}}{x}+1-\frac{1}{2x}\right)=\lim _{x\to \infty \:}\left(\frac{\sqrt{2x-3}}{x}\right)+\lim _{x\to \infty \:}\left(1\right)-\lim _{x\to \infty \:}\left(\frac{1}{2x}\right)\ (lim _{x\to \infty \:}\left(\frac{\sqrt{2x-3}}{x}\right) =\lim _{x\to \infty \:}\left(\frac{\frac{1}{\left(2x-3\right)^{\frac{1}{2}}}}{1}\right)=\lim _{x\to \infty \:}\left(\frac{1}{\left(2x-3\right)^{\frac{1}{2}}}\right)=\frac{\lim _{x\to \infty \:}\left(1\right)}{\lim _{x\to \infty \:}\left(\left(2x-3\right)^{\frac{1}{2}}\right)}=\frac{1}{\infty \:}=0 ;\lim _{x\to \infty \:}\left(1\right)=1;\lim _{x\to \infty \:}\left(\frac{1}{2x}\right)=\frac{1}{2}\cdot \lim _{x\to \infty \:}\left(\frac{1}{x}\right)=\frac{1}{2}\cdot \lim _{x\to \infty \:}\left(\frac{1}{x}\right)=\frac{1}{2}\cdot \frac{1}{\infty \:}=0 . \lim _{x\to \infty \:}\left(\frac{\sqrt{2x-3}}{x}+1-\frac{1}{2x}\right)=0+1-0=1)$

$\displaystyle\lim_{n \to -\infty} \dfrac{2x^3+2}{\sqrt{2x-3}+x-\dfrac{1}{2}}\\ ĐKXĐ:\left\{\begin{array}{l} 2x-3\ge 0\\\sqrt{2x-3}+x-\dfrac{1}{2} \ne 0\end{array} \right. \\\Leftrightarrow  x \ge \dfrac{3}{2}(Do\, x \ge \dfrac{3}{2} \Rightarrow \sqrt{2x-3}+x-\dfrac{1}{2} \ge 1 \ne 0 \, \forall \, x)\\ \\ \Rightarrow \not\exists \displaystyle\lim_{n \to -\infty} \dfrac{2x^3+2}{\sqrt{2x-3}+x-\dfrac{1}{2}}$