We have to find the limit `lim_(x->oo)sqrt(4x^2 + 5x) - 2x`

substituting `x = oo` gives the form `oo - oo` which is indeterminate but we cannot use l'Hopital's rule as it is not of the form `0/0 or oo/oo`

`lim_(x->oo)sqrt(4x^2 + 5x) - 2x`

`sqrt(4x^2 + 5x) - 2x`...

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We have to find the limit `lim_(x->oo)sqrt(4x^2 + 5x) - 2x`

substituting `x = oo` gives the form `oo - oo` which is indeterminate but we cannot use l'Hopital's rule as it is not of the form `0/0 or oo/oo`

`lim_(x->oo)sqrt(4x^2 + 5x) - 2x`

`sqrt(4x^2 + 5x) - 2x` = `(sqrt(4x^2+5x)-2x*sqrt(4x^2-5x)+2x)/(sqrt(4x^2 + 5x) + 2x)`

=> `((sqrt (4x^2 + 5x))^2 - (2x)^2)/(sqrt(4x^2 + 5x) + 2x)`

=> `(4x^2 + 5x - 4x^2)/(sqrt(4x^2 + 5x) + 2x)`

=> `(5x)/(sqrt(4x^2 + 5x) + 2x)`

let y = 1/x. As x tends to infinity, y tends to 0.

This changes the limit to

=> `lim_(y->0)(5/y)/(sqrt(4/y^2 + 5/y) + 2/y)`

=> `lim_(y->0)(5)/(sqrt(4+5y)+2)`

substitute y = 0

=> `5/(sqrt(4 + 0) + 2)`

=> 5/4

**The required limit is 5/4**

You must replace x by infinte and the limit yields oo-oo, which is impossible, therefore you must multiply and divide by the conjugate:

lim [sqrt(4x^2+5x)-2x][sqrt(4x^2+5x)+2x]/[sqrt(4x^2+5x)+2x]

Use the difference of two squares

(a-b)(a+b)=a^2-b^2

[sqrt(4x^2+5x)-2x][sqrt(4x^2+5x)+2x]=4x^2+5x-(2x)^2

[sqrt(4x^2+5x)-2x][sqrt(4x^2+5x)+2x]=4x^2+5x-4x^2

[sqrt(4x^2+5x)-2x][sqrt(4x^2+5x)+2x]=5x

lim [sqrt(4x^2+5x)-2x][sqrt(4x^2+5x)+2x]/[sqrt(4x^2+5x)+2x]=limit 5x/[sqrt(4x^2+5x)+2x]

Force the factor x^2 under the square root

limit 5x/{sqrt [x^2(4+5x/x^2)]+2x}=limit 5x/x{sqrt [(4+5x/x^2)]+2}

Reduce x:

limit 5/{sqrt [(4+5/x)]+2} =limit 5/[sqrt(4+5/infinite)+2]

5/infinite=0

limit 5/[sqrt(4+0)+2] = 5/[(sqrt4)+2]=5/4

**ANSWER: When x tends to infinite, the limit of the function is 5/4.**