# Proving by epsilon-delta that $\lim\limits_{t\to -1} (e^{t+1},|t+1|) = (1,0)$

Prove by epsilon-delta that the $\lim\limits_{t\to -1} \vec f(t)$ for $\vec{f}(t) = (e^{t+1},|t+1|)$ is $(1,0)$.

As much as I have progress, in order to get an upper bound by the definition and later propose a delta that depend of epsilon. I realized that $\sqrt{(e^{t+1}-1)^2+|t+1|^2} < \sqrt{(e^{t+1})^2+|t+1|^2}$, but I don't know what other function to use. Other than $x^x$ or $x!$, which usually increment faster than $e^x$, but they wouldn't help. Another thing that I'm trying to use, but I don't know if it would be useful, is that $e^x>x>\ln(x), \forall x \in \Bbb R$. Any ideas or suggestions?

## 1 Answer

You can treat each component separately and combine the results.

Let $\epsilon>0$ be given. Find a $\delta_1>0$ such that $t<\delta_2$ implies $\mid e^{t+1}-1 \mid <\epsilon/2$. Find a $\delta_2>0$ such that $t<\delta_2$ implies $\mid t+1 \mid <\epsilon/2$. Let $\delta = \min\{\delta_1, \delta_2\}$. Then for $t<\delta$, $$\sqrt{ (e^{t+1}-1)^2+\mid t+1 \mid^2 } < \sqrt{\frac{\epsilon^2}{4} + \frac{\epsilon^2}{4} } < \frac{\epsilon}{\sqrt{2}} < \epsilon.$$

• I know that could be an answer, but the fact is that my exercise tells me to do it by definition of a limit in $\Bbb R^n$, so I should use that expression and find an upper bound for it, but I haven't seen how? – Mounice Oct 5 '16 at 3:34
• @Mounice I believe that IS the definition of limit in $\mathbb{R}^2$. – B. Goddard Oct 5 '16 at 3:55