Take the 2-minute tour ×
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.

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)

share|improve this question
    
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
add comment

2 Answers

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., $0$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

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.