# Can't understand how to compute this limit where x tends to infinity

## Homework Statement

This is a question from a past exam paper:

Compute the limit:

lim x→∞
(2x$^{3}$ + x)$/$(3x$^{2}$ − 4x$^{3}$)

## The Attempt at a Solution

I really had no idea how to approach this but the solution is:
lim x→∞
(2x$^{3}$ + x)$/$(3x$^{2}$ − 4x$^{3}$)
= lim y→0+
(2 + y$^{2}$)$/$(3y − 4)
=
2$/$−4
= −1$/$2
Hopefully someone can explain to me the method used to obtain this answer.

## The Attempt at a Solution

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HallsofIvy
Homework Helper
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 $(2x^3+ x)/(3x^2- 4x^3)$, they have divided both numerator and denominator by the highest power of x, $x^3$, to get $(2+ 1/x^2)/(3/x- 4)$ 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.

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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.

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 $(2x^3+ x)/(3x^2- 4x^3)$, they have divided both numerator and denominator by the highest power of x, $x^3$, to get $(2+ 1/x^2)/(3/x- 4)$ 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.
Thanks very much for that - very clear.

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.
Thank you

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