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**RaisinBread** If you look at the first function, and pose $\displaystyle x_n = \frac{1}{(\frac{n \pi}{2})}$, well you can easily prove that $\displaystyle x_n$ converges to 0, since it is only the harmonic sequence multiplied by a constant, and you notice that $\displaystyle f(x_n)=sin((\frac{n \pi}{2}))$={1, 0, -1, 0, 1 ...}

It can easily be shown that f(xn) diverges, since there exists an epsilon greater than zero(just pick smaller than 1) for which, no matter what rank N of the sequence $\displaystyle f(x_n)$ you pick, there exists i greater than N such that $\displaystyle f(x_i)$-$\displaystyle f(x_{i+1})$ is greater than epsilon(two successive terms in f(xn) always have a distance of exactly 1), therefore $\displaystyle f(x_n)$ is not Cauchy, which implies that it does not converge.