What is the limit of # sqrt[x^3-4x^2+1] + sqrt[x^2-x] +x# as x goes to infinity?

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Answer 1 · 2015-09-17T15:51:52

#lim_(x->oo) sqrt[x^3-4x^2+1] + sqrt[x^2-x] +x = oo#

#lim_(x->oo) sqrt[x^3-4x^2+1] = oo#

#lim_(x->oo) sqrt[x^2-x] = oo#

#lim_(x->oo) x = oo#

Since the limit of each term is #oo#, the limit of the sum is #oo#

#lim_(x->oo) sqrt[x^3-4x^2+1] + sqrt[x^2-x] +x = oo#

If you want to give justification for the first 2 limits, use

#sqrt[x^3-4x^2+1] = sqrt(x^3) sqrt(1-4/x+1/x^3)#

and #sqrt[x^2-x] = sqrt(x^2) sqrt(1-1/x)#.

In each case, as #x->oo# the first factor also #->oo# and the second goes to #1#