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Warning : it's slightly stupid for a question, but i just want to assure it.

Given question :

$$\lim_{z \to 2i} \frac{z^2+4}{z-2i}=4i$$

My proof :

$$\forall \varepsilon\gt 0,\, \exists \delta\gt 0, \ni \, 0\lt \left|z-2i \right|\lt \delta \Rightarrow \left|\frac{z^2+4}{z-2i} - 4i \right|\lt \varepsilon$$ Sketch work : For $z\ne 2i$

$\begin{align} \left|\frac{z^2+4}{z-2i} - 4i \right| &=\left|\frac{z^2-4iz-4}{z-2i}\right| \\ &=\left|\frac{(z-2i)(z-2i)}{z-2i}\right| \\ &=\left|z-2i \right| \\ &\lt \varepsilon \\ \text{Set } \left|z-2i\right|\lt \varepsilon \text{ to be equal to } \delta, \text{ we conclude that } \varepsilon = \delta \end{align}$

Formal Proof:

Given $\varepsilon \gt 0$, choose $\delta = \varepsilon$ such that for $z\ne 2i$ we have the following:

Assume $0\lt \left|z-2i\right|\lt \delta$ $\begin{align} \Rightarrow\left|\frac{z^2+4}{z-2i} - 4i\right| &=\left|\frac{z^2-4iz-4}{z-2i}\right| \\ &=\left|\frac{(z-2i)(z-2i)}{z-2i}\right| \\ &=|z-2i|\\ &\lt \delta \\ &= \varepsilon \end{align}$

It's weird to get $\delta = \varepsilon$ when usually i got $\delta$ is multiple of $\varepsilon$. Is there anything wrong with my proof? If it isn't, could i conclude that "if we have a rational function with the numerator is a quadratic function that has two identical roots and the denominator is its factor and the difference of $z$ and $z_0$ is equal to the denominator, then if we wanna proof this limit, we have always get $\delta = \varepsilon$" ?

If you don't understand what i mean on the setence above, here is another example that i've tried and i got $\delta = \varepsilon$

$$\lim_{z \to i} \frac{z^2+1}{z-i}=2i$$

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In your first example, roots of the numerator are $2i$ and $-2i$ and for second example, roots are $i$ and $-i$.

So correct formulation is:

"If we have a rational function with the numerator is a quadratic function that has two roots which are negatives of each other and the denominator is its factor and the difference of $z$ and $z_0$ is equal to the denominator, then if we wanna proof this limit, we have always get $\delta=\epsilon$."

To verify this solve general example, $$\lim_{z \to a} \frac{z^2- a^2}{z-a}= 2a$$ where $a$ is any complex number.

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  • $\begingroup$ Hmmm.. actually my explanation isn't good. I mean, if we already combine the function with L, and all terms satisfiy with what i'm talking before... sorry for the mistake. Anyway, out of it... is that mean you agree with me that in this case $\delta=\varepsilon$? $\endgroup$ – user516076 Sep 24 '19 at 8:36
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    $\begingroup$ Yes! To verify this fact solve general example given above. $\endgroup$ – Mayuresh L Sep 24 '19 at 14:06

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