How do you find the limit of #(3x +9) /sqrt (2x^2 +1)# as x approaches infinity?

1 Answer
May 22, 2016

#3/sqrt2#

Explanation:

Factor the largest-degreed term from the numerator and denominator.

#lim_(xrarroo)(3x+9)/sqrt(2x^2+1)=lim_(xrarroo)(x(3+9/x))/sqrt(x^2(2+1/x^2))#

Pulling the #x^2# out of the square root as just #x# (because we are only concerned with positive value of #x#), this becomes

#lim_(xrarroo)(x(3+9/x))/(xsqrt(2+1/x^2))=lim_(xrarroo)(3+9/x)/sqrt(2+1/x^2)#

Now, when we evaluate the limit as #x# approaches infinity, we see that #9/x# and #1/x^2# both go to #0#.

#=(3+0)/sqrt(2+0)=3/sqrt2#