Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I don't understand why both limits have to be equal in both directions. $$\lim_{ \Delta x \rightarrow 0} {\Delta u+i\Delta v \over \Delta x} |_{\Delta y = 0} = \lim_{\Delta y \rightarrow 0}{\Delta u+i\Delta v \over i\Delta y}|_{\Delta x = 0} $$

Shouldn't the limit shift in which direction $x ,y$ moves on?

share|cite|improve this question
up vote 1 down vote accepted

Think of an analogy with the existence of limit of a function of one real variable. You need both the left and right hand limit to exist and be equal to the value of the function at that point. This is because there are two directions of approaching the particular point. Now, in case of a function of one complex variable, you have directions of approach in $x$ and $y$ directions in the complex plane. So you need to consider both $x$ and $y$ limits.

share|cite|improve this answer

It should be mentioned that $x$ and $y$ are both real so in this sense the LHS denotes when you're approaching the point in a direction parallel to the $x$-axis, whereas the RHS is when you're approaching along the $y$-axis. This gives the necessary case of Cauchy-Riemann by comparing real and imaginary parts, i.e we require $\partial u/\partial x=\partial v/\partial y$ and $\partial u/\partial y=-\partial v/\partial x$.

share|cite|improve this answer
i just meant to ask why they should be equal in both directions? parallel to x-axis and parallel to y-axis – Monkey D. Luffy Aug 25 '12 at 6:04
I guess i had been visualizing 3-Dimentional surface instead of curve. – Monkey D. Luffy Aug 25 '12 at 6:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.