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

by DanB1993
Tags: compute, infinity, limit
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DanB1993
#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
#2
Jan3-13, 05:15 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,497
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
#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
#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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