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

Asked on by jlykins14

lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

`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` .

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