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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$\displaystyle \frac{{5x^2 - 17}}{{x^2 + 2}} = \frac{{5 - \frac{{17}}{{x^2 }}}}{{1 + \frac{2}{{x^2 }}}}$
Originally Posted by Plato $\displaystyle \frac{{5x^2 - 17}}{{x^2 + 2}} = \frac{{5 - \frac{{17}}{{x^2 }}}}{{1 + \frac{2}{{x^2 }}}}$ So looking at this problem I should be able to "zero" out the denominators and my answer should just be $\displaystyle \frac{5}{1}=5$? Am I understanding correctly?
That is correct.
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