How do you find #\lim _ { x \rightarrow \infty } \frac { x ^ { 2} + x + 1} { ( 3x + 2) ^ { 2} }#?

2 Answers
maganbhai P.
Apr 15, 2018

#1/9#

Explanation:

We know that,
#color(red)(lim_(nto00)1/n=0=>lim_(nto00)1/n^2=0#

Here,

#lim_(xto00)(x^2+x+1)/(3x+2)^2#

#lim_(xto00)(x^2+x+1)/(9x^2+12x+4)#

Dividing numerator and denominator by #x^2!=0#

#lim_(xto00) [(x^2/x^2+x/x^2+1/x^2)/(9x^2/x^2+12x/x^2+4/x^2)]#

#=(lim_(xto00)(1+1/x+1/x^2))/(lim_(xto00)(9+12/x+4/x^2)#

#=(1+0+0)/(9+0+0)#

#=1/9#

Jim G.
Apr 15, 2018

#1/9#

Explanation:

#(x^2+x+1)/((3x+2)^2)=(x^2+x+1)/(9x^2+12x+4)#

#"divide terms on numerator/denominator by the highest"#
#"power of x that is "x^2#

#=(x^2/x^2+x/x^2+1/x^2)/((9x^2)/x^2+(12x)/x^2+4/x^2)=(1+1/x+1/x^2)/(9+12/x+4/x^2)#

#rArrlim_(xtooo)(x^2+x+1)/((3x+2)^2)#

#=lim_(xtooo)(1+1/x+1/x^2)/(9+12/x+4/x^2)#

#=(1+0+0)/(9+0+0)=1/9#

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