Question

Answer is 1/2

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Answer 1

Answer is 1/2

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Answer 2

Answer:

$\huge{\boxed{\frac{1}{2}}}$

Step-by-step explanation:

$\lim_{n \to \infty} \frac{\sqrt[3]{4+3n+n^6}}{1+3n+2n^2}\\\\\implies \lim_{n \to \infty} \frac{\sqrt[3]{\frac{4+3n+n^6}{n^6}}}{\frac{1+3n+2n^2}{n^2}}$

$\implies \lim_{n \to \infty} \frac{\sqrt[3]{\frac{4}{n^6}+\frac{3n}{n^6}+1}}{\frac{1}{n^2}+\frac{3n}{n^2}+2}}\\\\\implies \lim_{n \to \infty} \frac{1+0+0}{2+0+0}\\\\\implies \frac{1}{2}$