Wednesday, 18 September 2013

Proof that √x does not tend to a limit as x approaches infinity




I am wanting to prove from the definition of a limit ( ∀ε>0 ∃K>0:∀x>K, |f(x)−l|<ε) that √x does not tend to a limit as x approaches infinity.



So far I have tried to find a value of K such that |f(x)−l|<ε holds, and was planing on picking an x bigger than K that would give f(x)>l, however I am having trouble finding a value for K.



Thanks for any suggestions


Answer



By definition, \lim_{x\to\infty}f(x)=L if \forall\epsilon\gt 0 there exists N\gt 0 such that x\gt N\implies |f(x)-L|\lt\epsilon



|f(x)-L|\lt\epsilon is equivalent to L-\epsilon\lt f(x)\lt L+\epsilon.




To show that the limit does not exist, we can show that for any L, there is a fixed \epsilon for which the inequality cannot hold.



Since we are talking about the square root we can assume L is positive. Let \epsilon = 1. Then it is sufficient to show that we can always find x such that \sqrt{x}\ge L+1.



Let x=max(N+1, (L+2)^2). Then x\gt N and \sqrt{x}=\sqrt{(L+2)^2}=L+2\gt L+1 so L cannot be the limit. Since this is true for all L, the square root does not converge to a limit as x approaches infinity.


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