Proving the Limit of a Reciprocal Function

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In summary, to prove that lim x->a 1/F(x) = 0 given lim x->a F(x) = infinity, we use the definition of a limit to show that for any positive number M, there exists a delta value such that when x is within a certain distance from a, the value of F(x) is greater than M. This helps us to prove the desired limit by showing that as x approaches a, 1/F(x) approaches 0.
  • #1
Bigo75
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Homework Statement


Prove: If the Lim as x goes to a of F(x)=infinity, then lim as x goes to a of 1/F(x)=0


Homework Equations



We have only gone over limits and continuity using delta epsilion proofs.
 
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  • #2
what have you tried? this is straightforward in the sense that it follows from the definitions

show us what you have tried
 
  • #3
Since the limit is equal to infinity I took an M greater than zero such that X>M implies |F(x)-L| is less than epsilon but I get confused with the one over F(x)=0
 
  • #4
am I on the right track? I am really confused!
 
  • #5
Bigo75 said:
Since the limit is equal to infinity I took an M greater than zero such that X>M implies |F(x)-L| is less than epsilon but I get confused with the one over F(x)=0

this is wrong, lim x->a F(x) = inf means for all M > 0 there is a d > 0 s.t. |x-a| < d implies |F(x)| > M. (you can actually drop the absolute value sign here but i'll leave it)
 
  • #6
ok I understand this definition but how does that help me get lim as x goes to a of 1/f(x)=0? This definition will help me prove another one I was having trouble with. thank you.
 

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