Homework Help

The limit of `[sqrt(9x^2+x)-3x]` as x approaches infinity 

user profile pic

jlykins14 | eNotes Newbie

Posted September 19, 2013 at 3:27 AM via web

dislike 1 like

The limit of `[sqrt(9x^2+x)-3x]`

as x approaches infinity 

1 Answer | Add Yours

user profile pic

mjripalda | High School Teacher | (Level 1) Senior Educator

Posted September 19, 2013 at 4:33 AM (Answer #1)

dislike 0 like

`lim_(x->oo) sqrt(9x^2+x)-3x`

First, express the function as a fraction with denominator 1.

`=lim_(x->oo) (sqrt(9x^2+x)-3x)/1`

Then, multiply the numerator and denominator by the conjugate of `sqrt(9x^2+x)-3x` .

`=lim_(x->oo) (sqrt(9x^2+x)-3x)/1*(sqrt(9x^2+x)+3x)/(sqrt(9x^2+x)+3x)`

`=lim_(x->oo) (9x^2+x-9x^2)/(sqrt(9x^2+x)+3x)`

`=lim_(x->oo)x/sqrt((9x^2+x)+3x)`

Since the highest exponent of x present in the function is 1, multiply the numerator and denominator by the reciprocal of x.

`=lim_(x->oo)x/(sqrt(9x^2+x)+3x)*(1/x)/(1/x)`

Since `x=sqrt(x^2)` , the fraction in at the bottom simplifies to:

`=lim_(x->oo)1/(sqrt((9x^2+x)/x^2)+3)`

`= lim_(x->oo)1/(sqrt(9+1/x)+3)`

To take the limit, apply the property `lim_(x->oo)1/x^n=0` .

`= 1/(sqrt(9+0)+3)`

`=1/(sqrt9+3)`

`=1/(3+3)`

`=1/6`

Hence, `lim_(x->oo)sqrt(9x^2+x)-3x=1/6` .

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes