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If we directly evaluate the limit $\lim_{y\to\infty }\left(\frac{2y^2-3y+5}{y^2-5y+2}\right)$ as $y$ tends to $\infty $, we can see that it gives us an indeterminate form
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$\frac{\infty }{\infty }$
Learn how to solve equations différentielles problems step by step online. (y)->(infinity)lim((2y^2-3y+5)/(y^2-5y+2)). If we directly evaluate the limit \lim_{y\to\infty }\left(\frac{2y^2-3y+5}{y^2-5y+2}\right) as y tends to \infty , we can see that it gives us an indeterminate form. We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately. After deriving both the numerator and denominator, and simplifying, the limit results in. If we directly evaluate the limit \lim_{y\to\infty }\left(\frac{4y-3}{2y-5}\right) as y tends to \infty , we can see that it gives us an indeterminate form.
