# Interchanging limits

1. Jan 25, 2013

### EV33

1. The problem statement, all variables and given/known data

I was trying to prove something and I ended up in a situation similar to,

(limit t$\rightarrow$0)(limit s$\rightarrow$0) f(x+s,y+t)

=(limit s$\rightarrow$0)(limit t$\rightarrow$0)f(x+s,y+t)

My question is when does this equality hold. I can't find it anywhere?

Thank you.

2. Jan 25, 2013

### EV33

Thank you so much. My function is continuous.

3. Jan 25, 2013

### Dick

Sorry, deleted my answer because I was looking at something else and wanted to concentrate on that for a bit. But sure, if f is continuous at (x,y) you can interchange the limits. I was trying to think of a case where it's not true.

4. Jan 26, 2013

### HallsofIvy

Staff Emeritus
There exist examples, in most Calculus texts, which I don't have available now, of functions f(x, y) in which approaching (0, 0) along any straight line (such as going from (x, y) to (x, 0) then from (x, 0) to (0, 0), which is the same as "lim_(x->0)lim_(y->0) f(x, y)" or going from (x, y) to (0, y) then from (0, y) to (0, 0), which is the same as "lim_(y->0)lim_(x->0) f(x,y)) gives the same answer, the value of the function, so that "situation holds" but taking the limit along a quadratic curve gives a different answer so the function is NOT continuous.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}