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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.

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Are you trying to prove that directly from the definition? If so, what are you allowed to use about the exponential function? – Ricky Demer Aug 25 '11 at 1:53
@Ricky: Yes, using definition. Basic properties... what do you think? – Hiperion Aug 25 '11 at 2:02
Which basic properties, though? If you can use some fact like "e^y < 1 + y + y^2 for |y| < 1", it's quite easy! – Billy Aug 25 '11 at 2:29
You are right, it´s quite easy. thanks. – Hiperion Aug 25 '11 at 3:28
up vote 2 down vote accepted

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 $|y|\leq ||(x,y)||$.

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thank you very much! – Hiperion Aug 29 '11 at 0:46

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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easier :) +1 some kind of Lipschitzity arround $0$ – leo Aug 27 '11 at 23:25
thank you very much! – Hiperion Aug 29 '11 at 0:45

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