The question is to find this limit: $$\lim_{x\to\infty}\frac{2x^\frac{5}{3}- x ^\frac{1}{3}+7}{x^\frac{8}{5} +3x + \sqrt{x}}$$ I need any hint to help since I tried so much and couldn't solve it.
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3This, as it looks, is not a rational function. – Mikasa Nov 29 '13 at 14:46
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What is it called? – Mohammad Nov 29 '13 at 14:56
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Look at dominant coefficients of the numerator and denominator (those with highest degree), and factor them out, leaving $x$ to negative powers. Then, what you have factored out gives you the limit. – Jean-Claude Arbaut Nov 29 '13 at 15:10
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@user689 : I'm pretty sure the label "algebraic function" applies. Or you could just call it a "function". A rational function is the quotient of two polynomials. – Stefan Smith Nov 30 '13 at 02:37
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Hint: Divide both numerator and denominator with $x^{\alpha}$ where $$\alpha = \max \{\frac{5}{3}, \frac{1}{3},\frac{5}{8}, 1, \frac{1}{2}\}$$
and then take the limit
Belgi
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As $x$ tends to infinity, the dominant term of polynomial is the one with the largest power. Hence $$\lim_{x\to\infty}\frac{2x^\frac{5}{3}- x ^\frac{1}{3}+7}{x^\frac{8}{5} +3x + \sqrt{x}}=\lim_{x\to\infty}\frac{2x^\frac53}{x^\frac85}\to\infty$$
Shuchang
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I know this is only applicable to polynomials(i.e that contains only exponents belonging to the natural number set) right? – Mohammad Nov 29 '13 at 15:01
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since $\frac{5}{3}\gt\frac{8}{5}$, then the term with the largest degree is in the numerator. Thus your limit will tend toward $\infty$.
Eleven-Eleven
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$\lim_{x\to\infty}\frac{2x^\frac{5}{3}- x ^\frac{1}{3}+7}{x^\frac{8}{5} +3x + \sqrt{x}}=\lim_{x\to\infty}\frac{2x^\frac53}{x^\frac85}\to\infty$. If you want to do it in more steps divde the numerator and the denomerator by $x^{\frac{5}{3}}$, after this and taking the limit will show that the above equation is true.
user112167
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