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$\lim_{x\to\infty}\sqrt[3]{\frac{5+64x^5}{x^5+4}}$

Final answer to the exercise

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Other solved exercises

$\frac{m^6-64n^6}{m+3n}$

$-14x-12=12x+92$

$\frac{x^{21n}-1^{21n}}{x+1}\cdot$

$+\frac{2.7}{-3}$

$15x-9<6x-13$

$-2x^3-t^2+1=0$

$2\cdot0.1505$
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