# Limit by definition

Let $f(x,y)=\begin{cases}\dfrac{\mathrm{e}^{xy}-1}{x+y} & x\not=-y\; , \\ 0 & x=-y \; \end{cases}$ be a two variable function on $\mathbb{R}^2$.

How can I give a proof (Only by definition $\epsilon , \delta$) for$\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)=0$ ? (with details)

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Your expression is undefined in certain points arbitrarily close to $(0,0)$. Therefore the limit cannot exist. – Christian Blatter Dec 28 '12 at 10:10
Sorry, You are right. I fixed the problem. Now, Please help me. – bigli Dec 28 '12 at 10:37
In realy, I received to the problem through this excecise: prove that $h(x,y)=\mathrm{e}^{xy}$ is differentiable. (Only by the following definition) – bigli Dec 28 '12 at 14:06

The limit doesn't exist. Consider the sequence $$(x_n,y_n):=\left(-{1\over n},\ {1\over n}+{1\over n^3}\right)\qquad(n\geq1)\ .$$ As $$e^{xy}-1= x y\ g(x,y),\qquad \lim_{(x,y)\to(0,0)} g(x,y)=1\ ,$$ it follows that $$\lim_{n\to \infty}{e^{x_n y_n}-1\over x_n+y_n}=\lim_{n\to \infty} n^3\left(-{1\over n^2}+{1\over n^4}\right)=-\infty\ .$$ It's easy to produce another sequence $(x_n,y_n)$ where this limit is, e.g., $0$.
Evaluate the limit along the curve $y=-x+x^5$ as $x \to 0$: $$\lim_{x \to 0} \frac{e^{-x^2+x^6}-1}{x^5} = \lim_{x\to 0} \frac{-x^2}{x^5},$$ which does not exist.