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$$\lim_{n \rightarrow \infty } \frac 1n \cdot \frac {x^3}{1+x^2}$$

I think that for any fixed x the limit is $0$ due to increasing n. But I don't feel confident. Could you help?

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1 Answer 1

sure, that's true. if you fix any constant $c$ then you have $\lim_{n \rightarrow \infty} \frac{c}{n} = 0$ because for any $\epsilon > 0$ for all $n$'s larger than $\frac{|c|}{\epsilon}$ you have $|\frac{c}{n}| < |\frac{c}{\frac{c}{\epsilon}}| = \epsilon$

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