lim 6x^2+2x+1/5x^2-3x+4 x--> ∞ how do i solve the above questions ? can someone solve it and tell me and show me step by step please ? thank you
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$\displaystyle \lim_{x \to \infty} \frac{6x^2+2x+1}{5x^2-3x+4} = \lim_{x \to \infty} \frac{6+\frac{2}{x}+\frac{1}{x^2}}{5+ \frac{-3}{x}+\frac{4}{x^2}}\text{ ...that's from dividing top and bottom by } x^2.$
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