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

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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` .**

`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`