Proving Infinite Limit using Delta-Epsilon: One More Limit Homework Statement

  • Thread starter Thread starter Math_Geek
  • Start date Start date
  • Tags Tags
    Limit
Math_Geek
Messages
23
Reaction score
0

Homework Statement


lim as x goes to 1 from the right of 2^1/x-1=inf


Homework Equations



solve using delta-epsilon

The Attempt at a Solution



i am not sure how to prove an infinite limit, I have a defn that states, If for epsilon>0 there exists an M>0 such that x>M implies |f(x)-L|< epsilon. My main problem is that I am not sure how to do it, and how to get the power of two out of the way
 
Physics news on Phys.org
take the log
 
is the definition right?
 
did you mean lim x->1+ 2^(1/(1-x)) = 0?
 
Last edited:
no the problem says it goes to inf
 
Math_Geek said:
no the problem says it goes to inf

Oh it's lim x->1+ 2^(1/(x-1)), which is inf yea


The correct definition is

lim x->a+ f(x) = inf if for all M > 0 there is a d > 0 s.t. 0 < |x-1| < d and x > 1 implies |f(x)| > M
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top