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'Ey up!

I'm looking at limits of the form

$$ \lim_{x\to 0}\frac{\varphi(g(x))-\varphi(h(x))}{g(x)-h(x)} $$

where $\varphi$ is differentiable at $0$, and $g(x),h(x)\to 0$ as $x\to 0$. If $\varphi,g,h$ are "sufficiently nice", is this limit simply $\varphi'(0)$?

In particular, I am in the case where $\varphi$ is holomorphic and invertible, $f$ is holomorphic and has a fixed point at $\zeta$, and we have the limit

$$ \lim_{z\to\varphi(\zeta)}\frac{\varphi(f\circ\varphi^{-1}(z))-\varphi(\varphi^{-1}(z))}{f\circ\varphi^{-1}(z)-\varphi^{-1}(z)}. $$

Cauchy's integral formula might be of use.

N.N.
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1 Answers1

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I take the first half to be your question, and the second half to be motivation. Also, since $ g $ and $ h $ each have the limit $ 0 $ approaching $ 0 $ and their values at $ 0 $ are irrelevant, I'll assume that $ g ( 0 ) , \, h ( 0 ) = 0 $.

The answer to the first part is Yes if L'Hôpital's Rule applies: $$ \lim _ { x \to 0 } \frac { \varphi \bigl ( g ( x ) \bigr ) - \varphi \bigl ( h ( x ) \bigr ) } { g ( x ) - h ( x ) } \overset { ( 1 ) } = \frac { \varphi \bigl ( g ( 0 ) \bigr ) - \varphi \bigl ( h ( 0 ) \bigr ) } { g ( 0 ) - h ( 0 ) } = \frac { \varphi ( 0 ) - \varphi ( 0 ) } { 0 - 0 } = \frac 0 0 \text , $$ so $$ \lim _ { x \to 0 } \frac { \varphi \bigl ( g ( x ) \bigr ) - \varphi \bigl ( h ( x ) \bigr ) } { g ( x ) - h ( x ) } \overset { ( 2 ) } = \lim _ { x \to 0 } \frac { \varphi ' \bigl ( g ( x ) \bigr ) g ' ( x ) - \varphi ' \bigl ( h ( x ) \bigr ) h ' ( x ) } { g ' ( x ) - h ' ( x ) } = \frac { \varphi ' \bigl ( g ( 0 ) \bigr ) g ' ( 0 ) - \varphi ' \bigl ( h ( 0 ) \bigr ) h ' ( 0 ) } { g ' ( 0 ) - h ' ( 0 ) } = \frac { \varphi ' ( 0 ) \bigl ( g ' ( 0 ) - h ' ( 0 ) \bigr ) } { g ' ( 0 ) - h ' ( 0 ) } \overset { ( 3 ) } = \varphi ' ( 0 ) \text . $$

Now, the fine print justifying these calculations (see the numbers above some of the $ = $ signs) is:

  1. $ g $ and $ h $ must be continuous at $ 0 $, and then $ \varphi $ must be continuous at $ g ( 0 ) $ and $ h ( 0 ) $.
  2. $ g $ and $ h $ must be differentiable on a neighbourhood of $ 0 $, and then $ \varphi $ must be differentiable on neighbourhoods of $ g ( 0 ) $ and $ h ( 0 ) $.
  3. $ g ' ( 0 ) \ne h ' ( 0 ) $.

Since $ g ( x ) , \, h ( x ) = 0 $, (2) reduces to the requirement that all three functions must be differentiable on a neighbourhood of $ 0 $, and (1) is redundant. Also, if (3) fails, then we could try L'Hôpital's Rule again, and (although I'll skip the steps), it still works as long as all three functions are twice differentiable on a neighbourhood of $ 0 $ and $ g ' ' ( 0 ) \ne h ' ' ( 0 ) $. And if these second derivatives are equal, then we could apply L'Hôpital again, etc.

On the other hand, if $ g = h $ on a neighbourhood of $ 0 $, then your limit cannot exist. And if $ \phi $, $ g $, or $ h $ fails to be differentiable enough (if one of them isn't differentiable, if $ g ' ( 0 ) = h ' ( 0 ) $ and one of them isn't twice differentiable, etc, or if $ g ^ { ( n ) } ( 0 ) = h ^ { ( n ) } ( 0 ) $ for all $ n $ but $ g $ or $ h $ isn't analytic so that your limit still makes sense), then the argument breaks down, so your limit might not exist or might take the wrong value (although possibly a more sophisticated argument could rule this out).

Toby Bartels
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