What is the limit of #(3x^2+20x)/(4x^2+9)# as x goes to infinity?

Question

3510 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-10T12:41:40

#lim_(xrarroo)(3x^2+20x)/(4x^2+9)=3/4#

#(3x^2+20x)/(4x^2+9) = (x^2(3+20/x))/(x^2(4+9/x^2))# for #x != 0#

# = (3+20/x)/(4+9/x^2)# for #x != 0#.

As #xrarroo#, we get #20/x rarr0# and #9/x^2 rarr 0#, so we have

#lim_(xrarroo)(3x^2+20x)/(4x^2+9) = lim_(xrarroo)(3+20/x)/(4+9/x^2) =3/4#

Answer 2 · 2015-10-10T12:42:20

#3/4#

#lim_(x->oo) (3x^2+20x)/(4x^2+9) = lim_(x->oo) ((3x^2)/x^2+(20x)/x^2)/((4x^2)/x^2+9/x^2)= lim_(x->oo) (3+20/x)/(4+9/x^2)=3/4#

Note:

#20/x->0# when #x->oo#

#9/x^2->0# when #x->oo#