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How would you calculate this limit without use of derivatives? I know it goes to 1 but i can't seem to arrive at it.

$$\lim_{x\to\infty} \frac{\sqrt[\large4]{x^5} + \sqrt[\large 5]{x^3} + \sqrt[\large6]{x^8}}{\sqrt[\large 3]{x^4 + 2}}$$

amWhy
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Zerg Overmind
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  • You claim to know the limit. How have you come to know it? Was it given as a solution, provided in your text, to the question you ask? – amWhy Mar 23 '17 at 17:30
  • Knowing the limit means nothing if you don't know how to obtain it. – amWhy Mar 23 '17 at 17:32
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    Both the numerator and the denominator are $(1+o(1)),x^{4/3}$ for large values of $x$, hence the limit is clearly $\color{red}{1}$. – Jack D'Aurizio Mar 23 '17 at 17:34
  • Eh, that is exactly why i am asking here. I was given the limit and it's result but i'm having trouble figuring it out. – Zerg Overmind Mar 23 '17 at 17:38

3 Answers3

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HINT: note that $\sqrt[6]{x^8} = \sqrt[3]{x^4} = x^{4/3}$, and that $\frac{4}{3} > \frac{5}{4}$ and $\frac{4}{3} > \frac{3}{5}$.

Michael Seifert
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we write this as $$\frac{x^{5/4-4/3}+x^{3/5-4/3}+1}{\left(1+\frac{2}{x^4}\right)^{1/3}}$$ can you finish this?

Dr. Sonnhard Graubner
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Since $x \rightarrow \infty$ , $x^4+2 \approx x^4$

$$\lim_{x\to\infty} \frac{\sqrt[4]{x^5} + \sqrt[5]{x^3} + \sqrt[6]{x^8}}{\sqrt[3]{x^4 + 2}}=\lim_{x\to\infty} \frac{\sqrt[4]{x^5} + \sqrt[5]{x^3} + \sqrt[6]{x^8}}{\sqrt[3]{x^4 }} =\lim_{x\to\infty} \Bigg(\frac{x^{5/4}}{x^{4/3}} +\frac{x^{3/5}}{x^{4/3}}+\frac{x^{6/8}}{x^{4/3}}\Bigg) =\lim_ {x \to \infty}\Bigg( \frac{1}{x^{1/12}}+ \frac{1}{x^{11/15}}+1 \Bigg)=1$$

Jaideep Khare
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