I would like to prove that $$\lim_{\left( u, v \right) \rightarrow \left( 0, 0 \right)} \frac{u}{\sqrt{v+4}} = 0$$ using the $\epsilon - \delta$ definition of a limit.

My attempt is as follows:

Suppose $$0< \sqrt{u^2+v^2} < \delta $$

then $$\left| \frac{u}{\sqrt{v+4}} \right|=\frac{\left| u \right| }{\sqrt{v+4}}<\frac{\delta}{\sqrt{v+4}}.$$

If $\sqrt{v+4} \geq 1$, we have $$\frac{\delta}{\sqrt{v+4}} \leq\delta.$$

Therefore, $\delta = \epsilon$ should work.

I know that near the origin $\sqrt{v+4} \geq 1$. However, I would like a more sophisticated/complicated answer; for example, something like $$\delta = \min \left( \epsilon, c \right)$$ where $c>0$.

I would appreciate it if you point me in the right direction or provide a reference that deals with a similar problem.

  • $\begingroup$ Both answers were very helpful; as I can't designate both as the correct answer, I will designate the older one as such (though it's only a 2-minute difference). $\endgroup$ – Ahmed Ali Mar 5 at 20:50

Setting $\delta = \min\{\varepsilon, 1\}$ will work.

If $0 < \|(u,v)\| < \delta$, we have $|u|,|v| < \delta$ so $$\sqrt{v+4} \ge \sqrt{-|v|+4} > \sqrt{-\delta+4} \ge \sqrt{-1+4} = \sqrt{3}$$ and therefore $$\frac{|u|}{\sqrt{v+4}} < \frac{\delta}{\sqrt{3}} \le \frac{\varepsilon}{\sqrt3} < \varepsilon$$


First of all, please don't put "complicated" on a pedestal! Good mathematics strives to explain complicated things as simply as possible. The end goal is simplicity, that as many people as possible can understand. In mathematics research, if you've taken something that nobody understands and turned it into something only you can understand, it is barely an improvement!

That said, your solution is incomplete, and by completing it, you'll get a $\delta$ in the form you want. As you say,

I know that near the origin $\sqrt{v + 4} \ge 1.$

You'll need to encode this into your $\delta$. If you just choose $\delta = \varepsilon$, then having $\varepsilon = 3.99$ would cause the inequality to fail. If $\varepsilon > 4$, then $0 < \sqrt{u^2 + v^2} < \varepsilon$ will not guarantee $(u, v)$ lies in the domain of the function!

To correct this, let's consider where $\sqrt{v + 4} \ge 1$. Note that this is equivalent to $v \ge -3$. If we force $\delta \le 3$, then this should work, as $$0 < \sqrt{u^2 + v^2} < 3 \implies v > -3.$$ Therefore, just choose $\delta = \min \{\varepsilon, 3\}$.

  • 1
    $\begingroup$ I am wholly convinced by your words. I think that my wording has betrayed me. What I wanted to convey is that I don't want to depend on just the intuition that $\sqrt{v+4} \geq 1$ near the origin; I wanted a more concrete/solid proof. Thank you for your answer. $\endgroup$ – Ahmed Ali Mar 5 at 20:58

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