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$\lim _{x \rightarrow \infty}\left(\frac{x^{2}}{3 x-2}-\frac{x}{3}\right)=$
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Verified Answer
The correct answer is:
$\frac{2}{9}$
Consider $\lim _{x \rightarrow \infty}\left[\frac{x^{2}}{3 x-2}-\frac{x}{3}\right]$
$$
\begin{array}{l}
=\lim _{x \rightarrow \infty}\left[\frac{3 x^{2}-x(3 x-2)}{3(3 x-2)}\right] \\
=\lim _{x \rightarrow \infty} \frac{2 x}{3(3 x-2)}=\lim _{x \rightarrow \infty} \frac{2 x}{3 x\left[3-\frac{2}{x}\right]} \\
=\lim _{x \rightarrow \infty} \frac{2}{3} \frac{1}{\left(3-\frac{2}{x}\right)}=\frac{2}{3} \times \frac{1}{3-0}=\frac{2}{9}
\end{array}
$$
$$
\begin{array}{l}
=\lim _{x \rightarrow \infty}\left[\frac{3 x^{2}-x(3 x-2)}{3(3 x-2)}\right] \\
=\lim _{x \rightarrow \infty} \frac{2 x}{3(3 x-2)}=\lim _{x \rightarrow \infty} \frac{2 x}{3 x\left[3-\frac{2}{x}\right]} \\
=\lim _{x \rightarrow \infty} \frac{2}{3} \frac{1}{\left(3-\frac{2}{x}\right)}=\frac{2}{3} \times \frac{1}{3-0}=\frac{2}{9}
\end{array}
$$
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