Find the limit... ?

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

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Answer 1 · 2017-02-14T23:55:11

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

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:

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