How do you find the limit of #(3x^2 - 5x + 1) / (4x^2 + 5x)# as x approaches infinity?

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Answer 1 · 2016-09-11T15:51:52

By substitution, we find that #lim_(x->oo)(3x^2-5x+1)/(4x^2+5x)=oo/oo#

Using l'hopital's rule:

#lim_(x->oo)(d/dx(3x^2-5x+1))/(d/dx(4x^2+5x))#

#lim_(x->oo)(6x-5)/(8x+5)=oo/oo#

l'hopital's rule again:

#lim_(x->oo)(d/dx(6x-5))/(d/dx(8x+5))#

#lim_(x->oo)6/8=3/4#