Wednesday, November 29, 2017

real analysis - textFindlimnrightarrowinftyfracnan



Find  limnnan where a > 0 is a number.






To solve this problem we first realize that we need to use l'hopitals rulebecause if we evaluate at infinity we get 



limnnan=using l'hopitals rule we get limn1nxn1=0If we take lhopitals rule again we get limn0(n1)nxn2=0Thus   limnnan=0Is this the correct procedure that one uses in order to find the limit for this problem?Is there any mistakes in my solution?



Answer



L'Hospital's rule is only applicable when you have a limit of the form





or



00



Thus we need to consider 3 cases: (a) $0, (b) a=1, and (c) a>1.




(a) For a<1 we have



limnnan=



because limnan=0 and limnn=, so we have



$$\frac{\infty}{0}=\infty, \quad 0



(b) When a=1, we have




limnnan=limnn1n=limnn=,a=1



(c) For a>1 we have



limnnan=



This is the only case where L'Hospital's rule is applicable. Here, write



limnnan=limnddn(nan)=limnddn(nexp(ln(an)))=limnddn(nexp(nln(a)))=limnddn(1ln(a)exp(nln(a)))=limnddn(1ln(a)an)=1limnnan=0,a>1



I added extra steps to demonstrate how to perform the derivative of an.


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