What is the limit of #(2x+3)/(5x+7)# as x goes to infinity?

2 Answers
Oct 4, 2015

#lim(x->oo)(2x+3)/(5x+7)=2/5#

Explanation:

#lim(x->oo)(2x+3)/(5x+7)#

Notice that the degree of the numerator and the denominator are

the same i.e: 1, for this and all the similar scenarios the limits is

simply the ratio of the leading coefficients of top to bottom:

#:.lim(x->oo)(2x+3)/(5x+7)=2/5#

Oct 4, 2015

As #x->oo#, the #3# and #7# become insignificant relative to the magnitude of #x#. In other words, #x# becomes so large that:

#lim_(x->oo) (2x + 3)/(5x + 7) = lim_(x->oo) (2color(red)(cancel(color(black)(x))))/(5color(red)(cancel(color(black)(x))))#

#= lim_(x->oo) 2/5#

#= color(blue)(2/5)#