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