# Evaluate the limit : $$\displaystyle{\lim_{x \to \infty}}$$ $$x^2 + x + 1\over {3x^2 + 2x – 5}$$

## 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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