I'm trying to prove this limit: $\lim_{(x,y)\to (0,0)}(3xy+1,e^y+2)=(1,3)$ I know the definition, but can't bound the norm. Thanks for your help.
Limit in two variables
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0You are right, it´s quite easy. tha$n$ks. – 2011-08-25
2 Answers
In order to reduce the number of unanswered questions, I answer this:
Let $\epsilon\gt 0$. You know that $\lim_{y\to 0} e^y = 1,$ so, exist \delta'>0 such that $|e^y-1|<\frac{\epsilon}{2}.$ Take \delta=\min\left \{ \delta',\sqrt{\frac{\epsilon}{6}}\right\}. Thus, if $||(x,y)||<\delta$, $\begin{align*} ||(3xy+1,e^y+2)-(1,3)||&\leq |3xy| + |e^y-1|\\ &\leq 3||(x,y)||^2+|e^y-1|\\ &< 3 \cdot \frac{\epsilon}{6} + \frac{\epsilon}{2}\\ &=\epsilon. \end{align*}$ The last inequality follows because $|y|\leq ||(x,y)||$.
Let $u(x,y)=(3xy+1,e^y+2)$, hence $u(0,0)=(1,3)$. Since one is interested in the limit of $u$ at $(0,0)$, one can (and we will) assume that $\|(x,y)\|\le1$ where $\|\ \|$ denotes the Euclidean norm.
Using $|xy|\le |y|$ and $|e^y-1|\le2|y|$ for every $(x,y)$ such that $\|(x,y)\|\le1$, and $|y|\le\|(x,y)\|$ for every $(x,y)$, one gets $ |u(x,y)-u(0,0)|\le5\|(x,y)\|. $
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0easier :) +1 some kind of Lipschitzity arround $0$ – 2011-08-27