<s>[INST] Find $\lim\limits_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$ $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$$
I'm surprisingly struggling with this limit, could you give me a hint how to handle it (no L'Hospital and no prior knowledge what the limit value is)? 
Generally what are some basic methods to handle $\infty \cdot 0$? 
 [/INST] $$\lim _{ n\rightarrow \infty  } n\left( \cos  \left( \frac { 1 }{ \sqrt { n }  }  \right) -1 \right) =\lim _{ n\rightarrow \infty  } n\left( \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) - }  } \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) - }  }  \right) =\\ =\lim _{ n\rightarrow \infty  } n\left( -2\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right)  }  \right) =-\frac { 1 }{ 2 } \lim _{ n\rightarrow \infty  }{ \left( \frac { \sin { \left( \frac { 1 }{ 2\sqrt { n }  }  \right)  }  }{ \frac { 1 }{ 2\sqrt { n }  }  }  \right)  } ^{ 2 }=-\frac { 1 }{ 2 } $$
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