The problem involves computing a limit as x approaches infinity for the expression (2x^{3} + x)/(3x^{2} − 4x^{3}). This falls under the subject area of calculus, specifically limits and asymptotic behavior.
Several participants have provided insights into the methods used to approach the limit, including the technique of simplifying the expression by focusing on the highest power terms. There is an ongoing exploration of different perspectives on the problem, but no explicit consensus has been reached.
Participants note the challenge of dealing with limits at infinity and the common practice of transforming the expression to make it more manageable. There is an implicit understanding of the constraints imposed by the homework context, but specific rules or assumptions are not detailed.
Compute the limit:
lim x→∞
(2x[itex]^{3}[/itex] + x)[itex]/[/itex](3x[itex]^{2}[/itex] − 4x[itex]^{3}[/itex])
lim x→∞
(2x[itex]^{3}[/itex] + x)[itex]/[/itex](3x[itex]^{2}[/itex] − 4x[itex]^{3}[/itex])
= lim y→0+
(2 + y[itex]^{2}[/itex])[itex]/[/itex](3y − 4)
=
2[itex]/[/itex]−4
= −1[itex]/[/itex]2
HallsofIvy said:It's a fairly standard technique- "infinity" is hard to deal with while "0" is very easy- so they have made use of the fact if x goes to infinity, 1/x goes to 0.
Starting from [itex](2x^3+ x)/(3x^2- 4x^3)[/itex], they have divided both numerator and denominator by the highest power of x, [itex]x^3[/itex], to get [itex](2+ 1/x^2)/(3/x- 4)[/itex] and then replaced "1/x" with "y": (2+y^2)/(3y- 4). Then, as x goes to infinity, y goes to 0 so that the limit is just (2+ 0)/(0- 4)= 2/(-4)= -1/2.
mtayab1994 said:Or at infinity you take the highest power in the numerator over the highest power in the denominator. In your case it would be 2x^3/-4x^3 and then x^3 will cancel out leaving you with -2/4 which is -1/2.