Friday, 4 April 2014

calculus - Finding a limit with squeeze theorem



Let an=1n2+1+1n2+2....1n2+n+1




Find limnan.



I know the limit is zero. I also know that I need to use the squeeze theorem to solve it.



Now using nn2+1annn2+n+1 as my upper and lower bounds doesn't yield 0, it yields 1.



It doesn't seem right but can I just use 1n2+1an1n2+n+1 to solve ?



Thanks.



Answer



The limit is not 0, in fact you are almost there!



n+1n2+n+1ann+1n2+1



For LHS, limnn+1n2+n+1=limn1+1n1+1n+1n2=1.



For RHS, limnn+1n2+1=limn1+1n1+1n2=1



Hence, by Squeeze Theorem, limnan=1



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