Approaching the Limit: Solving a Tricky Exam Problem

  • Thread starter thepatient
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    Exam Limit
In summary, the problem was about finding the limit of (x*y^2)/(x^2+y^4) as (x,y) approaches (0,0). The initial attempt was to use paths and then the squeeze theorem to determine the limit, with the final conclusion being the limit is 0. However, there was discussion about potential errors in the approach and it was later determined that the limit does not exist. The student received partial credit for their attempt.
  • #1
thepatient
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limit problem from past exam. :(

Homework Statement


Problem from an exam that I just did is really bugging me.

lim(x,y)->(0,0) (x*y2)/(x2+y4)
The attempt at a solution[/b]
At first I thought the limit didn't exist. I tried using paths, x=y1/2, but then I couldn't find a counter example for which a path approached at a different point. So then I tried using the squeeze theorem, and got that the limit is 0. :\

Was my approach to the problem correct? is the limit really 0? [
 
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  • #2


You didn't say how you proved it. But yes, the limit is zero.
 
  • #3


Dick said:
You didn't say how you proved it. But yes, the limit is zero.
Oh really? XD Well I assumed:

0<y2<x2+y4 (since x^2 and y^4 are always positive, and y2<y4

0<y2/x2+y4<1
Then multiplying x into the inequality:

0<x*y2/x2+y4<x

Then taking the limit:
lim(x,y)->(0,0) 0<lim(x,y)->(0,0) x*y2/x2+y4<lim(x,y)->(0,0) x

Which gave me that:
0<lim(x,y)->(0,0) x*y2/x2+y4<0

which by the squeeze theorem, limit is zero. Was that correct...?
 
  • #4


thepatient said:
Oh really? XD Well I assumed:

0<y2<x2+y4

0<y2/x2+y4<1
Then multiplying x into the inequality:

0<x*y2/x2+y4<x

Then taking the limit:
lim(x,y)->(0,0) 0<lim(x,y)->(0,0) x*y2/x2+y4<lim(x,y)->(0,0) x

Which gave me that:
0<lim(x,y)->(0,0) x*y2/x2+y4<0

which by the squeeze theorem, limit is zero. Was that correct...?

Well the starting point 0<y^2<x^2+y^4 is wrong. Suppose x=0 and y=(1/2). I would use polar coordinates.
 
  • #5


Limit doesn't exist , try y=x^(1/2)
 
  • #6


Aaah... I was starting it with polar coordinates at first and it seemed like the best way to go, but then it seemed to messy so I thought I was doing it wrong. I was running out of time and just left it like that. XD Thanks lots, at least maybe I have partial credit haha..
 
  • #7


lim x->0 f(x,x^(1/2))=1/2
and lim x->0 f(x,0)=0
 
  • #8


tt2348 said:
Limit doesn't exist , try y=x^(1/2)

Ooops. You are so right. Sorry.
 
  • #9


Aww darn... it was a 3 part question worth 5 points total on that part of the exam, so I should have 10/3 credit hehe... The rest of the test was pretty good.
 

Related to Approaching the Limit: Solving a Tricky Exam Problem

What is a limit problem?

A limit problem is a mathematical problem that involves finding the value that a function approaches as the input approaches a certain value. It is used to describe the behavior of a function near a particular point.

How do I solve a limit problem?

To solve a limit problem, you can use various techniques such as direct substitution, factoring, and algebraic manipulation. You can also use graphing or numerical methods to approximate the limit.

What is the importance of limit problems?

Limit problems are important because they help us understand the behavior of a function and its values at specific points. They also have many applications in real-world scenarios, such as in physics, engineering, and economics.

What are the common types of limit problems?

The most common types of limit problems include finding a limit at a specific point, evaluating one-sided limits, and determining limits at infinity. Other types include finding limits of trigonometric, exponential, and logarithmic functions.

What are some tips for solving limit problems?

Some tips for solving limit problems include identifying the type of limit problem, using algebraic techniques to simplify the expression, and checking for any potential indeterminate forms. It is also helpful to use a graph or table to visualize the behavior of the function near the given point.

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