
#1
Jan313, 05:05 AM

P: 6

1. The problem statement, all variables and given/known data
This is a question from a past exam paper: 3. The attempt at a solution I really had no idea how to approach this but the solution is: 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Jan313, 05:15 AM

Math
Emeritus
Sci Advisor
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. 



#3
Jan313, 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.




#4
Jan313, 08:13 AM

P: 6

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


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