What is #lim_(ntooo)sqrt(n^2+n^a)−n# , while a is real number?

Ive tried to multiply it as #(a-b)(a+b)# so I get to #(n^a)/(sqrt(n^2+n^a)+n)#
But for #a=2,a<2,a>2# I cant get the right limit.

1 Answer
Andrea S.
Apr 11, 2018

#lim_(n->oo) sqrt(n^2+n^a)-n = {(0 " when " a < 1), (1/2 " when " a = 1), (oo " when "a > 1):}#

Explanation:

Multiply and divide by:

#sqrt(n^2+n^a)+n#

to have:

#sqrt(n^2+n^a)-n = ((sqrt(n^2+n^a)-n)(sqrt(n^2+n^a)+n))/(sqrt(n^2+n^a)+n)#

and using the algebraic identity #(a-b)(a+b) = a^2-b^2#:

#sqrt(n^2+n^a)-n = (n^2+n^a-n^2)/(sqrt(n^2+n^a)+n)#

#sqrt(n^2+n^a)-n = n^a/(sqrt(n^2+n^a)+n)#

Now divide and multiply by #1/n#:

#sqrt(n^2+n^a)-n = (n^a/n)/((sqrt(n^2+n^a)+n)/n)#

#sqrt(n^2+n^a)-n =n^(a-1)/(sqrt(1+n^(a-2))+1#

So, for #a > 1# we have:

#lim_(n->oo) n^(a-1) = oo#

and:

#lim_(n->oo) sqrt(n^2+n^a)-n = oo#

For #a =1#

we have:

#n^(a-1) = 1#

#lim_(n->oo) n^(a-2) =lim_(n->oo) 1/n = 0#

#lim_(n->oo) sqrt(n^2+n^a)-n = 1/2#

For #a < 1#

we have:

#lim_(n->oo) n^(a-1) = 0#

#lim_(n->oo) sqrt(n^2+n^a)-n = 0#

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