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Say we are to evaluate the following complex-valued limit $(1)$ via the $\varepsilon$$-$$\delta$ definition.

$$\lim\limits_{z\to i}\left(\frac{iz^3 - 1}{z + i}\right), \ \ \ \ z \in \Bbb{C},\tag1$$


I have tried different techniques, such as factorisation, multiplying by complex conjugate, various inequalities... Still, I always hit a dead end.

  • How can I prove that this limit is $0$ via the epsilon$-$delta definition?
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  • $\begingroup$ Since this is a seeded question, I am happy to accept an answer that provides a different epsilon$-$delta proof (shorter, explains in a different way, includes graphs...). As always, a bit scared that my proof is incorrect, so if anyone drops a comment, it shall be corrected. :=) $\endgroup$ Mar 3, 2017 at 21:58
  • $\begingroup$ You did notice that you can just evaluate at $z = i$ (or do you mean $\lim_{z \to -i}$, or $z - i$ in the denominator)? $\endgroup$ Mar 4, 2017 at 9:53
  • $\begingroup$ @Magdiragdag, of course; this is just a useful case to see in terms of the epsilon-delta proof technique (in my opionion). More of a useful pathway to follow when in need to apply the definition of a limit. :) $\endgroup$ Mar 4, 2017 at 9:59

1 Answer 1

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Let us first rearrage $(1)$ a bit to a nicer form for our purposes.

$$\lim\limits_{z\to i}\left(\frac{iz^3 - 1}{z + i}\right) = \lim\limits_{z\to i}\left[\frac{i\left(z^3 - 1/i\right)}{z + i}\right] = \lim\limits_{z\to i}\left[\frac{i\left(z^3 + i\right)}{z + i}\right]\tag{1'}$$

First find an estimate of what the limit is. From $(1')$ with direct substitution,

$$\lim\limits_{z\to i}\left[\frac{i\left(z^3 - 1/i\right)}{z + i}\right] = \frac{i\left(-i + i\right)}{2i} = 0. \tag2$$

What remains is to prove that $0$ actually is the limit of $(1)$.

$$\left|\frac{i\left(z^3 + i\right)}{z + i} - 0\right| = \left|\frac{z^3 + i}{z + i}\right| = \left|\frac{(z - i)\left(z^2 + iz - 1\right)}{z + i}\right| \leq \left|\frac{|z - i|\left(|z|^2 + |z| + 1\right)}{z + i}\right|\tag 3$$

$$\frac{|z - i|\left(|z|^2 + |z| + 1\right)}{|z + i|} < \delta\frac{|z|^2 + |z| + 1}{|z + i|}\tag{3'}$$

Choose $\delta < 1$; this guarantees an open punctured unit disc centered at $i$. In this region, $|z| < 2$ from which

$$|z|^2 + |z| + 1 < 7.$$

In contrast, we need a lower bound for $|z + i|$. This is a bit trickier. Notice that in this open region given by $\delta < 1$ all complex numbers have a positive imaginary part. That is, $\Im{(z)} = b > 0$.

$$|z + i| = |a + bi + i| = |a + (b + 1)i| = \sqrt{a^2 + (b + 1)^2} > 1\tag4$$

In conclusion,

$$0 < |z - i| < \delta \implies \left|\frac{i\left(z^3 + i\right)}{z + i} - 0\right| < 7\delta < \varepsilon.\tag5$$

Choosing

$$\delta = \min\left\{1,\frac{\varepsilon}{7}\right\}\tag6$$

completes the proof.

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