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student:mathematics:limits [2018/08/26 14:27] – bernstdhstudent:mathematics:limits [2024/01/24 13:34] – bernstdh
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- +===== Limits =====
-===== Mathematical Foundations: Limits =====+
  
  
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 Congratulations!  You've just taken your first limit.  How would you write this down?  There are two ways.  One way is to write: if \(n \rightarrow  \infty \) then \(1/n \rightarrow 0\) (i.e., if \(n\) goes to infinity then \(1/n\) goes to zero). Another way is to write: Congratulations!  You've just taken your first limit.  How would you write this down?  There are two ways.  One way is to write: if \(n \rightarrow  \infty \) then \(1/n \rightarrow 0\) (i.e., if \(n\) goes to infinity then \(1/n\) goes to zero). Another way is to write:
  
-|   \(lim_{n \rightarrow  \oo} 1/n = 0\)   |+|   \(lim_{n \rightarrow  \infty} 1/n = 0\)   |
  
 which says that the limit of \(1/n\) as \(n\) goes to infinity is zero. which says that the limit of \(1/n\) as \(n\) goes to infinity is zero.
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   - Given two functions of \(x\), \(f_{1}\) and  \(f_{2}\), with finite limits, \(L_{1}\) and \(L_{2}\) it follows that \(lim_{x \rightarrow N}(f_{1} + f_{2}) = L_{1} + L_{2}\)   - Given two functions of \(x\), \(f_{1}\) and  \(f_{2}\), with finite limits, \(L_{1}\) and \(L_{2}\) it follows that \(lim_{x \rightarrow N}(f_{1} + f_{2}) = L_{1} + L_{2}\)
   - Given two functions of \(x\), \(f_{1}\) and  \(f_{2}\), with finite limits, \(L_{1}\) and \(L_{2}\) it follows that then \(lim_{x \rightarrow N}(f_{1} f_{2}) = L_{1} L_{2}\)   - Given two functions of \(x\), \(f_{1}\) and  \(f_{2}\), with finite limits, \(L_{1}\) and \(L_{2}\) it follows that then \(lim_{x \rightarrow N}(f_{1} f_{2}) = L_{1} L_{2}\)
-  - (L'Hopital's Rule) Given two differentiable functions of \(x\),  +  - (L'Hopital's Rule) Given two differentiable functions of \(x\), \(f_{1}\) and  \(f_{2}\), with infinite limits and derivatives \(f_{1}^{\prime}\) and \(f_{2}^{\prime}\), it follows that \(lim_{x \rightarrow \infty}(\frac{f_{1}}{f_{2}}) =  lim_{x \rightarrow \infty}(\frac{f_{1}^{\prime}}{f_{2}^{\prime}})\). 
-\(f_{1}\) and  \(f_{2}\), with infinite limits and derivatives \(f_{1}\prime\) and \(f_{2}\prime\), it follows that \(lim_{x \rightarrow \oo}(\frac{f_{1}}{f_{2}}) =  lim_{x \rightarrow \oo}(\frac{f_{1}\prime}{f_{2}\prime})\). +