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$\displaystyle{}\underset{n \to \infty}{\lim} \frac{1}{n} \bigg( \frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n} \bigg)$ is equal to

  1. $\infty$
  2. $0$
  3. $\log_e 2$
  4. $1$
in Calculus
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Is $0$ is the answer?

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