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We have to find limit of $f(z)=(z-2)\log |z-2|$ at $z_0=2$

My try: This is in $0^0$ indeterminate form.Can I use L'Hôpital's rule here$?$.Thank you.

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    How is this $0^0$? – 2017-02-14
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    $f(z)=\log (|z-2|)^{(z-2)}$.correct me if i'm wrong. – 2017-02-14
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    You are wrong, because $a\cdot \log(b)=\log(b^a)\neq (\log(b))^a$. – 2017-02-14
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    Is $z$ real or complex? – 2017-02-14
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    $z$ is complex number – 2017-02-14
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    @5xum:i also wrote the same. – 2017-02-14
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    @MatheMagic So then you have $\log(0^0)$, not just $0^0$. – 2017-02-14
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    By writing $$ (z-2)\ln |z-2|=\frac{\ln |z-2|}{1/(z-2)} $$ one gets, as $z \to 2$, an undeterminate form $-\frac{\infty}\infty$ then one is allowed to use L'Hospital's rule as shown below. – 2017-02-14

2 Answers 2

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Let $\ln |u|$ be the natural logarithm defined over all $u\neq0$.

Assume first that $z$ is a real number.

One may use L'Hospital's rule by setting $u=z-2$ and as $z \to 2$ getting $u \to 0$ and $$ \lim_{z\to 2}(z-2)\ln |z-2|=\lim_{u\to0}u\ln |u|=\lim_{u\to0}\frac{\ln |u|}{1/u}=\lim_{u\to0}\frac{\dfrac1u}{-\dfrac1{u^2}} \:=\:-\lim_{u\to 0}u = 0. $$ Now if $z \in \mathbb{C}$, as $z \to 2$, one can't apply L'Hospital's rule to $z \mapsto (z-2)\log |z-2|$ ($\log$ being any branch cut of the logarithm) since $z \mapsto z-2$ is holomorphic but $z \mapsto \log |z-2|$ is not.

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    $\ln|u|$ is real so can we say $\dfrac{d}{dz}\ln x=\dfrac{1}{z}$.? – 2017-02-14
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    @MyGlasses Answer edited. Thank you. – 2017-02-14
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L'Hospital's rule is not the alpha and omega of limits computation. Actually, if $z$ is a real number, you can use the standard high-school limit: $$\lim_{x\to 0}x\ln\lvert x\rvert=0$$ and set $u=\lvert z-2\rvert$, since for any function $f$ of the complex variable $z$, $$\lim f(z)=0\iff\lim\lvert f(z)\rvert=0.$$.