How do you determine the limit of #sqrt(2x^2 +3x) - 4x# as x approaches #oo#?

Question

William Abdalla · Accepted Answer

We have that

#sqrt(2x^2 +3x) - 4x=(sqrt(2x^2 +3x) - 4x)*(sqrt(2x^2 +3x) +4x)/(sqrt(2x^2 +3x) + 4x)= [(sqrt(2x^2+3x))^2-(4x)^2]/((sqrt(2x^2 +3x) + 4x))= (2x^2+3x-16x^2)/((sqrt(2x^2 +3x) + 4x))= (3x-14x^2)/[x(sqrt(2+3/x)+4))= x^2(3/x-14)/[x(sqrt(2+3/x)+4))= x(3/x-14)/[sqrt(2+3/x)+4] #

Hence the limit is

#lim_(x->oo) x*[3/x-14]/[sqrt(2+3/x)+4] =(oo) times [0-14]/(sqrt2+4)=-oo#

Footnote
WolframAlpha agrees with the above result as well

Scarlett Allison · Answer

undefined To determine the limit of sqrt(2x^2 + 3x) - 4x as x approaches infinity, we can simplify the expression by dividing every term by the highest power of x in the expression, which is x^2. By doing this, we get:

sqrt(2x^2 + 3x) / x^2 - 4x / x^2

Simplifying further, we have:

sqrt(2 + 3/x) - 4

As x approaches infinity, the term 3/x approaches zero. Therefore, the expression simplifies to:

sqrt(2) - 4

Hence, the limit of sqrt(2x^2 + 3x) - 4x as x approaches infinity is sqrt(2) - 4.