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


by DanB1993
Tags: compute, infinity, limit
DanB1993
DanB1993 is offline
#1
Jan3-13, 05:05 AM
P: 6
1. The problem statement, all variables and given/known data

This is a question from a past exam paper:

Compute the limit:

lim x→∞
(2x[itex]^{3}[/itex] + x)[itex]/[/itex](3x[itex]^{2}[/itex] − 4x[itex]^{3}[/itex])
2. Relevant equations



3. The attempt at a solution

I really had no idea how to approach this but the solution is:
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
Hopefully someone can explain to me the method used to obtain this answer.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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HallsofIvy
HallsofIvy is offline
#2
Jan3-13, 05:15 AM
Math
Emeritus
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Thanks
PF Gold
P: 38,877
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
mtayab1994 is offline
#3
Jan3-13, 07:10 AM
P: 584
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.

DanB1993
DanB1993 is offline
#4
Jan3-13, 08:13 AM
P: 6

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


Quote Quote by HallsofIvy View Post
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.
Thanks very much for that - very clear.

Quote Quote by mtayab1994 View Post
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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