1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Multivariable limit

  1. Oct 30, 2014 #1
    1. The problem statement, all variables and given/known data
    find lim as x,y approach 0 of (10sin(x^2 + y^2)) / (x^2 + y^2)

    2. Relevant equations


    3. The attempt at a solution
    direct substitution yields indeterminate form and so does multiplying by the conjugate. what other methods are there to use?
     
  2. jcsd
  3. Oct 30, 2014 #2

    Zondrina

    User Avatar
    Homework Helper

    So you want to find:

    $$\displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{10sin(x^2 + y^2)}{x^2 + y^2}$$

    Try approaching ##(0, 0)## along different paths to zero such as ##(0, x)## and ##(x, 0)##.

    EDIT: Path ##(t, t)## looks promising.
     
  4. Oct 30, 2014 #3
    for the paths (0,x) and (x,0) I got 10sin(y^2) / y^2. does that mean the limit is 10sin(y^2) / y^2
     
  5. Oct 30, 2014 #4

    Zondrina

    User Avatar
    Homework Helper

    What is:

    $$\displaystyle \lim_{x \rightarrow 0} \frac{10sin(x^2)}{x^2}$$

    Looks like a first year problem.
     
  6. Oct 30, 2014 #5
    wow. cant believe I didn't see that . so that is indeterminate but I know the limit exists because the problem says "Hint - the limit does exist". when you said the path (t,t) looked promising I assumed you meant parameterizing x and y. but what do I let them equal
     
  7. Oct 30, 2014 #6

    Zondrina

    User Avatar
    Homework Helper

    Indeed you can approach ##(0, 0)## along several different paths and get the same answer. That's how you know the limit exists and is finite. Plugging in ##x = t## and ##y = t## will give the same limit.
     
  8. Oct 30, 2014 #7
    so the paths (x,0) (0,x) (y,0) (0,y) and (t,t) all give indeterminate forms. Im just not seeing a way around this
     
  9. Oct 30, 2014 #8

    Zondrina

    User Avatar
    Homework Helper

    Why not apply L'Hospital's rule? If the form is ##0/0## you can easily find the limit that way.

    Although the conventional way would be to recognize: ##\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1##
     
  10. Oct 30, 2014 #9
    I was told l'hospital's rule did not apply in three dimensions. but since one variable goes away when evaluated on (x,0) , (y,0), etc... does that mean its like that variable never existed
     
  11. Oct 30, 2014 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No matter how ##(x,y) \to (0,0)## the distance of ##(x,y)## from the origin goes to 0. In other words, the squared distance ##r^2 = x^2 + y^2 \to 0##. In fact, ##r^2 \to 0## if, and only if ##(x,y) \to (0,0)## in some way. Now go back and re-examine your original function ##f(x,y)##.
     
  12. Oct 31, 2014 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes. The limits you end up with using those paths only depend on one variable, so you can use the Hospital rule on them. Unfortunately, all you'll have shown is that the limit along those paths exist. You actually need to show the original limit exists for all possible paths to the origin. You want to think about what Ray said.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Multivariable limit
  1. Multivariable limits (Replies: 3)

  2. Multivariable Limit (Replies: 4)

  3. Limit in Multivariable (Replies: 3)

  4. Multivariate Limit (Replies: 3)

  5. Multivariable Limit (Replies: 13)

Loading...