I have to evaluate the following limit:

$$\lim_{z\rightarrow \frac{\sqrt3i}{2}}\frac{z}{4z^2+3}$$

I know the limit doesn't exist. I tried proving it by letting: $$z = x +iy,\\y =\frac{\sqrt3i}{2} $$

And then evaluating the new limit for $$x\rightarrow0\_$$

and $$x\rightarrow0_+$$

and hence show they do not equal one another. However, this approach didn't work since I got to $$\lim_{x\rightarrow0\_}\frac{x+i\sqrt3/2}{4x(x+i\sqrt3)}$$

and got stuck.

Could I get some help guys?

Edit: sorry if the formatting isn't that good. I'm still new to this site.

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    $\begingroup$ The expression has the form $c/0$ where $c\ne 0$. (And your development is wrong.) $\endgroup$ – Yves Daoust May 1 '17 at 19:46
  • $\begingroup$ @YvesDaoust So, if it's simply c/0, it's enough to say the limit doesn't exist without any other proof? $\endgroup$ – Abdul Miah May 1 '17 at 19:52
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    $\begingroup$ Yep, you are at a pole. $\endgroup$ – Yves Daoust May 1 '17 at 19:53
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    $\begingroup$ First thing when you meet a limit is to evaluate the function ! $\endgroup$ – Yves Daoust May 1 '17 at 19:55
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    $\begingroup$ Sorry, your development wasn't wrong, my bad. You are not stuck, when $x\to0$, your expression $\to\infty$. $\endgroup$ – Yves Daoust May 1 '17 at 19:58

We can represent the expression as \begin{align*} \lim_{z\rightarrow \frac{\sqrt3i}{2}}\frac{z}{4z^2+3} =\lim_{z\rightarrow \frac{\sqrt{3}i}{2}}\frac{z}{4\left(z+\frac{\sqrt{3}i}{2}\right)\left(z-\frac{\sqrt{3}i}{2}\right)} \end{align*}

We observe $\frac{\sqrt{3}i}{2}$ is a pole of first order of $\frac{z}{4z^2+3}$ and so the limit does not exist.


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