What is $\lim_{x\to\infty} \frac{\sqrt[4]{x^5} + \sqrt[5]{x^3} + \sqrt[6]{x^8}}{\sqrt[3]{x^4 + 2}}$?

Question

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.

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

Answer 1

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?

Answer 2

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

Answer 3

Since $x \rightarrow \infty$ , $x^4+2 \approx x^4$

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