Exercise 5.10

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Exercise 10: Suppose $f$ and $g$ are complex differentiable functions on $(0,1)$, $f(x)ightarrow 0$, $g(x)ightarrow 0$, $f'(x)ightarrow A$, $g'(x)ightarrow B$
    as $xightarrow 0$, where $A$ and $B$ are complex numbers, $B
eq 0$. Prove that $$\lim_{xightarrow 0}\frac{f(x)}{g(x)}=\frac{A}{B}.$$ Compare with Example 5.18.

analambanomenos · Accepted Answer

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Let $f(x)=u(x)+iv(x)$ where $u,v$ are real-valued and differentiable on $(0,1)$, and let $A=a+ib$. Following the hint, since $$\frac{u'(x)}{1}\rightarrow a\quad
    \hbox{and}\quad\frac{v'(x)}{1}\rightarrow b\quad\hbox{as $x\rightarrow 0$,}$$ we have by Theorem 5.13 $$\frac{u(x)}{x}\rightarrow a\quad\hbox{and}\quad
    \frac{v(x)}{x}\rightarrow b\quad \hbox{as $x\rightarrow 0$}$$ so that $$\frac{f(x)}{x}\rightarrow A\quad\hbox{as $x\rightarrow 0$}.$$ Similarly, by breaking $g$ into
    its real and imaginary parts, we get $$\frac{g(x)}{x}\rightarrow B\quad\hbox{and so}\quad\frac{x}{g(x)}\rightarrow\frac{1}{B}\quad\hbox{as $x\rightarrow 0$}.$$ Hence
    $$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0} \Biggl(\biggl(\frac{f(x)}{x}-A\biggr)\frac{x}{g(x)}+A\frac{x}{g(x)}\Biggr)=\frac{A}{B}.$$

In Example 5.18 it was shown that the derivative of the denominator $g$ satisfies $\big|g'(x)\big|\ge (2/x)-1$. Since the limit of $g'(x)$ as $x\rightarrow 0$ does
    not exist, we can't apply the above result to this case.

Exercise 5.10

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Exercise 5.10

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