# solve algebraically (without the use of derivatives): `lim_(x->+oo)sqrt(4x^2+3x)-2x`

sciencesolve | Certified Educator

You need to use the following algorithm to evaluate the limit without the use of derivatives, hence, you need to start by multiplying by the conjugate of the expression `sqrt(4x^2 + 3x) - 2x ` such that:

`lim_(x->oo) (sqrt(4x^2 + 3x) - 2x) = lim_(x->oo) ((sqrt(4x^2 + 3x) - 2x)(sqrt(4x^2 + 3x) + 2x))/(sqrt(4x^2 + 3x) + 2x)`

Converting the product `((sqrt(4x^2 + 3x) - 2x)(sqrt(4x^2 + 3x) + 2x))` into a difference of squares, yields:

`lim_(x->oo) (4x^2 + 3x - 4x^2)/(sqrt(4x^2 + 3x) + 2x)`

Reducing duplicate members to numerator yields:

`lim_(x->oo) (3x)/(sqrt(4x^2 + 3x) + 2x)`

You need to force factor out `4x^2` under the square root, such that:

`lim_(x->oo) (3x)/(sqrt(4x^2(1 + 3x/(4x^2))) + 2x) `

`lim_(x->oo) (3x)/(|2x|sqrt(1 + 3/(4x)) + 2x)`

Since `x->oo` , hence `|2x| = 2x.`

Factoring out `2x` yields:

`lim_(x->oo) (3x)/(2x(sqrt(1 + 3/(4x)) + 1))`

Reducing duplicate factors yields:

`lim_(x->oo) (3)/(2(sqrt(1 + 3/(4x)) + 1)) = (3/2)*lim_(x->oo) 1/(sqrt(1 + 3/(4x)) + 1)`

Using the properties of limits, yields:

`(3/2)*lim_(x->oo) 1/(sqrt(1 + 3/(4x)) + 1) = (3/2)*1/(sqrt(1 + lim_(x->oo) 3/(4x)) + 1)`

`lim_(x->oo) (3x)/(2x(sqrt(1 + 3/(4x)) + 1))= (3/2)*1/(sqrt(1+0) + 1)`

`lim_(x->oo) (3x)/(2x(sqrt(1 + 3/(4x)) + 1)) = (3/2)*(1/2)`

Hence, evaluating the given limit, without the use of derivatives, yields `lim_(x->oo) (sqrt(4x^2 + 3x) - 2x) = 3/4` .

aruv | Student

`y(x)=sqrt(4x^2+3x)-2x`

we want to find `lim_(x->+oo)y(x)=lim_(x->+oo)(sqrt(4x^2+3x)-2x)`

Multiply and divide the expression by `sqrt(4x^2+3x)+2x`  ie.

`lim_(x->+oo)y(x)=lim_(x->+oo){(sqrt(4x^2+3x)-2x)(sqrt(4x^2+3x)+2x)}/(sqrt(4x^2+3x)+2x)`

`=lim_(x->+oo)(4x^2+3x-4x^2)/(sqrt(4x^2+3x)+2x)`

`=lim_(x->+oo)(3x)/{xsqrt(4+3/x)+2x}`

`=lim_(x->+oo)(3x)/{x(sqrt(4+3/x)+2)}`

`=lim_(x->+oo)3/(sqrt(4+3/x)+2)`

`=3/(sqrt(4)+2)`

`=3/(2+2)`

`=3/4`

`lim_(x->+oo)(sqrt(4x^2+3x)-2x)=3/4`