If $\vec{a}$ is defined like this: $$\vec{a_k}=\binom{\frac{1}{k+1}\cos{k}}{\;\;(k+1)\sin{(\frac{1}{k+1})}\;\;}$$ Does it then converge? And if so, what is the limit? Any help on how to approach this problem would be very helpful.
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Sorry, I'm studying in german and got a bit confused. Changed now – Void Jul 04 '16 at 22:00
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Yes. The expression $\frac{1}{k+1}$cos$(k)$ converges to 0 as $k\rightarrow \infty $ since $\frac{1}{k+1}$ approaches zero and cos $(k)$ just oscillates between -1 and 1. The expression $(k+1)$sin$(\frac{1}{(k+1)})$ approaches 1 as $k\rightarrow \infty$ since
$(k+1)$sin$(\frac{1}{(k+1)}) \approx \frac{k+1}{k+1}$ for large k.
Since each of its components converge, the vector function also converges to $(0,1)$ as $k\rightarrow \infty $
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