# `lim_(x->+oo) sqrt(x^2+4x)-sqrt(x^2+6x)=?`what is the limit in plus infinity square root of x squared plus 4x minus square root of x squared plus 6x?Thank you!!

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### 1 Answer

`lim_(x->+oo) sqrt(x^2+4x) - sqrt(x^2+6x)`

If we substitute `x=+oo` to the function, it results to:

`lim_(x->+oo) sqrt(x^2+4x)-sqrt(x^2+6x)=oo-oo`

which is indeterminate.

So, to evaluate its limit, multiply the function by its conjugate.

`lim_(x->+oo) sqrt(x^2+4x)-sqrt(x^2+6x) * (sqrt(x^2+4x) + sqrt(x^2+6x))/(sqrt(x^2+4x)+sqrt(x^2+6x))`

`=lim_(x->+oo) (x^2+4x - (x^2+6x))/(sqrt(x^2+4x)+sqrt(x^2+6x))=lim_(x->+oo)(-2x)/(sqrt(x^2+4x)+sqrt(x^2+6x))`

To be able to divide each by the term with the highest degree which is x, multiply the top and bottom by 1/x.

`=lim_(x->+oo) (-2x)/(sqrt(x^2+4x)+sqrt(x^2+6x)) * (1/x)/(1/x)= lim_(x->+oo)((-2x)/x)/(sqrt(x^2+4x)/x+sqrt(x^2+6x)/x)`

Note that `x=sqrt(x^2)` .

`=lim_(x->+oo) ((-2x)/x)/(sqrt((x^2+4x)/x^2)+sqrt((x^2+6x)/x^2))= lim_(x->+oo) (-2)/(sqrt(1+4/x)+sqrt(1+6/x))`

Then, apply the property `lim_(x->+oo)1/x=0` .

`=(-2)/(sqrt(1+0)+sqrt(1+0))=(-2)/(sqrt1+sqrt2)=(-2)/2=-1`

**Hence, `lim_(x->+oo) sqrt(x^2+4x)-sqrt(x^2+6x) = -1.`**