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

In summary, the technique used to compute the limit in the given problem is to divide both the numerator and denominator by the highest power of x, replace 1/x with y, and then take the limit as y approaches 0. This allows for easier evaluation of the limit as x approaches infinity.
  • #1
DanB1993
6
0

Homework Statement



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])

Homework Equations





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.
 
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  • #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 [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.
 
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  • #3
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
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.

Thanks very much for that - very clear.

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.

Thank you
 
Last edited by a moderator:

1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept in calculus that describes the behavior of a function as its input value approaches a certain value, typically infinity or a specific number. It is used to determine the value that a function approaches as its input value gets closer and closer to a particular value.

2. How do you compute a limit?

To compute a limit, you must first determine the function's behavior as the input value approaches the given value. This can be done by evaluating the function at values that are increasingly closer to the given value. If the function approaches a finite value, then that value is the limit. If the function approaches infinity or negative infinity, then the limit does not exist.

3. What does it mean when a limit tends to infinity?

When a limit tends to infinity, it means that the function's output becomes larger and larger as the input value approaches infinity. This can be represented by the notation "lim f(x) = ∞" or "lim f(x) = -∞" for a function f(x).

4. What are some common techniques for computing limits?

Some common techniques for computing limits include direct substitution, factoring and canceling, using algebraic manipulations, and using L'Hopital's rule. It is important to also consider the domain of the function and any discontinuities when computing limits.

5. How do I know if a limit exists?

A limit exists if the function's output approaches a finite value as the input value approaches a specific value. This can be determined by evaluating the function at values that are increasingly closer to the given value and checking if the outputs form a pattern or approach a specific value. If the function approaches infinity or negative infinity, then the limit does not exist.

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