Answer
[tex]\text{The value is -}\frac{1}{3}[/tex]
Given
[tex]\lim _{x\to\infty}\frac{2x^2(x-3)}{3x^2(1-2x)}[/tex]
Solution
Let take the common factor first
[tex]\frac{2}{3}\lim _{x\to\infty}\frac{x^2(x-3)}{x^2(1-2x)}[/tex]
Let's apply algebraic property:
[tex]a+b=a(1+\frac{b}{a})[/tex][tex]\begin{gathered} \frac{2}{3}.\lim _{x\to\infty}\frac{x^2(x-3)}{x^2(1-2x)} \\ \\ \frac{2}{3}.\lim _{x\to\infty}\frac{x^2x(1-\frac{3}{x})}{x^2x(\frac{1}{x}-2)} \\ \\ x^2x\text{ can divide each other } \\ \\ \frac{2}{3}.\lim _{x\to\infty}\frac{(1-\frac{3}{x})}{(\frac{1}{x}-2)} \end{gathered}[/tex]
Let take the limit of the numerator
[tex]\begin{gathered} \lim _{x\to\infty}(1-\frac{3}{x})=1 \\ \end{gathered}[/tex]
Also take the limit of the denominator
[tex]\lim _{x\to\infty}(\frac{1}{x}-2)=-2[/tex]
[tex]we\text{ now have }\frac{1}{-2}[/tex]
Now, let not forget the common factor outside
[tex]\frac{2}{3}.\frac{1}{-2}=\frac{2}{-6}=-\frac{1}{3}[/tex]
So therefore, the Limit exist and the value is
[tex]-\frac{1}{3}[/tex]
