In this week's lecture we discussed L'Hopital's Rule and that it can be used to determine the type of singularity. Judging from the examples shown I assume that when after applying L'Hopital's Rule one gets 0, then it's a pole of some degree. If it's $\neq 0$, but still finite, then it's a removable singularity.

Is that really the case? Because I have been doing some workbook exercises and $\frac{sin(z)-tan(z)}{z^2}$ is supposed to have a removable singularity at z=0, and after applying L'Hopital's Rule twice I land on $\frac{-sin(z)-2sec^2(z)tan(z)}{2}$ which is clearly 0 for z=0, meaning it would be a pole of degree 2 (because I had to use L'Hopital's Rule twice).

Any help would be appreciated.

  • $\begingroup$ The function you are considering is of the form $f(z) = \frac{h(z)}{g(z)}$ with $g(z),h(z)$ analytic at $z=0$. Therefore there exists $k,n \in \mathbb{N}$ such that $g(z) = z^n G(z)$ and $h(z) = z^k H(z)$ with $G(0) = \frac{g^{(n)}(0)}{n!}\ne 0,H(0) =\frac{h^{(n)}(0)}{n!}\ne 0$, and $f(z)$ is analytic at $z=0$ iff $k\ge n$. In that case $f(0) = \lim_{z \to 0} f(z) = \lim_{z \to 0} z^{k-n} \frac{H(0)}{G(0)}$. $\endgroup$
    – reuns
    Commented Nov 21, 2016 at 8:45
  • $\begingroup$ It is a removable singularity, not a pole. Let $f(0)=0$ and let $f(z)=(\sin z-\tan z)/z^2$ for $z\ne 0.$ Then $f$ is analytic. $\endgroup$ Commented Nov 21, 2016 at 12:07
  • $\begingroup$ @user254665 Yes, but how do I see that from L'Hopital's Rule? There are problems where I get similarly 0 using L'Hopital's Rule but the singularity is a pole of some degree. So how do I see what singularity it is just by using L'Hopital's Rule. $\endgroup$
    – Rab
    Commented Nov 21, 2016 at 15:50

2 Answers 2


(I)...Let $U$ be a nbhd of $0.$ Let $f$ and $g$ be analytic and not constant on $U$ with $f(0)=g(0)=0,$ and let $g(z)\ne 0$ for $z\in U$ \ $\{0\}.$ For $z\in U$ we have $f(z)=z^af_1(z)$ and $g(z)=z^bg_1(z),$ where $a,b$ are positive integers, and $f_1,g_1$ are analytic on $U$ with $f_1(0)\ne 0\ne g_1(0).$

So $f(z)/g(z)=z^{a-b}f_1(z)/g_1(z)$ for $z\in U$ \ $\{0\}.$ Therefore we have

(i). If $a>b$ then $\lim_{z\to 0}f(z)/g(z)=0.$

(ii).If $a=b$ then $\lim_{z\to 0}f(z)/g(z)=f_1(0)/g_1(0)\ne 0.$

(iii).If $a<b$ then $\lim_{z\to 0}f(z)/g(z)$ does not exist.

In (i) and (ii) the point $0$ is a removable singularity of $f/g$ . In (iii), $f/g$ has a pole of order $b-a$ at $0.$

(II)...Let $h^{(0)}=h$ and let $h^{(j)}$ be the $j$-th derivative of $h$ for $j>0.$ Applying the rule $$(z^ch(z))^{(n)}=\sum_{j=0}^n\binom {n}{j}(z^c)^{(j)}(h(z))^{(n-j)}$$ with $h=f_1$ and with $h=g_1,$ we see that if you need l'Hopital $m$ times ($m>0$), but not less than $m$ times, to find that $\lim_{z\to 0} f(z)/g(z)$ exists, then $m=b=a$ if the limit is non-zero, while $m=b<a$ if the limit is $0.$

In particular, with $f(z)=\sin z - \tan z$ and $g(z)=z^2$ then for $z\ne 0$ we have $$f(z)/g(z)=(\;(z-z^3/6+...)-(z+z^3/3+...)\;)/z^2=(-z^3/4+... )/z^2.$$ And $f'(z)/g'(z)=(-3z^2/4+...)/2z.\quad$ And $f''(z)/g''(z)=(-3z+...)/2.$


Why do you complicate matters? L'Hospital's theorem is great for functions of a real variable, that may be differentiable without being analytic. For a complex variable, though, everything complex-differentiable is holomorphic, so use Taylor series to your advantage.

In your concrete case, $\sin' 0 = \cos 0 = 1$ and $\sin '' 0 = \cos ' 0 = -\sin 0 = 0$, which means that the Taylor series of $\sin$ around $0$ is $\sin z = z + z^3 f(z)$ with $f$ holomorphic around $0$.

Similarly, $\tan' 0 = \frac 1 {\cos^2 0} = 1$ and $\tan '' (0) = 2 \frac {\tan 0} {\cos^2 0} = 0$, so the Taylor series of $\tan$ around $0$ is $\tan z = z + z^3 g(z)$ with $g$ a holomorphic function around $0$.

It follows, then, that around $0$

$$\frac {\sin z - \tan z} {z^2} = \frac {z^3 (f(z) - g(z))} {z^2} = z (f(z) - g(z))$$

which is a holomorphic function, therefore $0$ is a removable singularity for that fraction.


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