Question

Find the limit... ?

Find the limit... ? please THANK YOU (っ＾▿＾)

Answer 1

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

Explanation:

Evaluate the limit:

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

The limit is in the form `oo-oo` that is indeterminate.

Intuitively we should note that `x^2` is an infinite of higher order and it should prevail: we can prove it however by separating that factor:

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

For `x->oo` we can assume `x` positive, so:

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

Now the quantity under the root has a finite limit:

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

so that the limit we need to evaluate is determinate:

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