Show that $\left\{{\dfrac{3n^{2}+5n-2}{6n^{2}-6n+5}}\right\}$ converges and find its limit.

Solution The function $$f(x)= \dfrac{3x^{2}+5x-2}{6x^{2}-6x+5} \quad x>0$$

is a related function of the sequence $\left\{ \dfrac{3n^{2}+5n-2}{6n^{2}-6n+5}\right\}$. Since $$\begin{eqnarray*} \lim\limits_{x\,\rightarrow \,\infty }\,f(x)&=&\lim_{ x \rightarrow \,\infty }\frac{3x^{2}+5x-2}{6x^{2}-6x+5} =\lim_{ x \rightarrow \,\infty }\frac{\dfrac{3x^{2}}{6x^{2}}+\dfrac{5x}{6x^{2}}-\dfrac{2}{6x^{2}}}{1-\dfrac{6x}{6x^{2}}+\dfrac{5}{6x^{2}}}\\[5pt] &=&\lim_{x\rightarrow \,\infty }\frac{\dfrac{1}{2}+\dfrac{5}{6x}-\dfrac{1}{3x^{2}}}{1-\dfrac{1}{x}+\dfrac{5}{6x^{2}}}=\dfrac{\dfrac{1}{2}+0-0}{1-0+0}=\dfrac{1}{2} \end{eqnarray*}$$

the sequence $\left\{ \dfrac{3n^{2}+5n-2}{6n^{2}-6n+5}\right\}$ converges and $\lim\limits_{n\,\rightarrow \,\infty } \dfrac{3n^{2}+5 n-2}{6n^{2}-6n+5}=\dfrac{1}{2}$.