Solution :
\(\displaystyle{\lim_{x \to \infty}}\)\(x^2 + x + 1\over {3x^2 + 2x – 5}\)
It is (\(\infty\over \infty\) form), Put x = \(1\over y\)
= \(\displaystyle{\lim_{y \to 0}}\) \(1 + y + y^2\over {3 + 2y – 5y^2}\) = \(1\over 3\)
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