I've seen the Limit set to zero to solve the equation, but here I have infinity. How should I approach it?

$\displaystyle \lim_{x\to infinity}\frac{5x^2-17}{x^2+2}$

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- Feb 19th 2008, 11:37 AMXIII13ThirteenLimits in equation --->Infinity question
I've seen the Limit set to zero to solve the equation, but here I have infinity. How should I approach it?

$\displaystyle \lim_{x\to infinity}\frac{5x^2-17}{x^2+2}$ - Feb 19th 2008, 11:56 AMPlato
$\displaystyle \frac{{5x^2 - 17}}{{x^2 + 2}} = \frac{{5 - \frac{{17}}{{x^2 }}}}{{1 + \frac{2}{{x^2 }}}}$

- Feb 20th 2008, 01:13 PMXIII13Thirteen
- Feb 20th 2008, 01:38 PMPlato
That is correct.