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Here is Prob. 10, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ and $g$ are complex differentiable functions on $(0, 1)$, $f(x) \to 0$, $g(x) \to 0$, $f^\prime(x) \to A$, $g^\prime(x) \to B$ as $x \to 0$, where $A$ and $B$ are complex numbers, $B \neq 0$. Prove that $$ \lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{A}{B}.$$ Compare with Example 5.18. Hint: $$ \frac{f(x)}{g(x)} = \left\{ \frac{f(x)}{x} - A \right\} \cdot \frac{x}{g(x)} + A \cdot \frac{x}{g(x)}.$$ Apply Theorem 5.13 to the real and imaginary parts of $\frac{f(x)}{x}$ and $\frac{g(x)}{x}$.

Now here is Theorem 5.13 in Baby Rudin, 3rd edition:

Suppose $f$ and $g$ are real and differentiable in $(a, b)$, and $g^\prime(x) \neq 0$ for all $x \in (a, b)$, where $-\infty \leq a < b \leq +\infty$. Suppose $$\frac{f^\prime(x)}{g^\prime(x)} \to A \ \mbox{ as } \ x \to a.$$ If $$ f(x) \to 0 \ \mbox{ and } \ g(x) \to 0 \ \mbox{ as } \ x \to a,$$ or if $$ \tag{15} g(x) \to +\infty \ \mbox{ as } \ x \to a,$$ then $$\frac{f(x)}{g(x)} \to A \ \mbox{ as } \ x \to a.$$ The analogous statement is of course also true if $x \to b$, or if $g(x) \to -\infty$ in (15). . . .

And, here is Example 5.18:

On the segment $(0, 1)$, define $f(x) = x$ and $$g(x) = x + x^2 \mathrm{e}^{\iota/x^2}. \tag{35}$$ Since $\left| \mathrm{e}^{\iota t} \right| = 1$ for all real $t$, we see that $$\lim_{x \to 0} \frac{f(x)}{g(x)} = 1. \tag{36} $$ Next, $$g^\prime(x) = 1+ \left\{ 2x - \frac{2\iota}{x} \right\} \mathrm{e}^{\iota/x^2} \qquad \qquad (0 < x < 1), \tag{37} $$ so that $$ \left| g^\prime(x) \right| \geq \left| 2x- \frac{2\iota}{x} \right| - 1 \geq \frac{2}{x} - 1. \tag{38} $$ Hence $$ \left| \frac{f^\prime(x)}{g^\prime(x)} \right| = \frac{1}{\left| g^\prime(x) \right|} \leq \frac{x}{2-x} \tag{39} $$ and so $$ \lim_{x\to 0} \frac{f^\prime(x)}{g^\prime(x)} = 0. \tag{40} $$ By (36) and (40), L'Hospital's rule fails in this case. Note also that $g^\prime(x) \neq 0$ on $(0, 1)$, by (38).

I think I fully understand the calculation in Example 5.18.

The difference between the situation in Example 5.18 and the one in Prob. 5 is that in the former case $\lim_{x\to 0} g^\prime(x)$ fails to exist, whereas in the latter not only do both $\lim_{x\to 0} f^\prime(x)$ and $\lim_{x\to 0} g^\prime(x)$ exist but $\lim_{x\to 0} g^\prime(x)$ is non-zero as well. Hence the difference between the resulting conclusions. Am I right?

My Attempt at a solution to Prob. %, Chap. 5:

We can prove the folliwng: If $h$ is a complex function defined on an interval $[a, b]$ on the real line and if $p \in [a, b]$, then $$\lim_{x \to p} h(x) = A,$$ where $A$ is a complex number, if and only if $\lim_{x \to p} \Re h(x)$ and $\lim_{x \to p} \Im h(x)$ both exist and equal $\Re A$ and $\Im A$, respectively. Moreover, $h$ is differentiable at $p$ if and only if both $\Re h$ and $\Im h$ are differentiable at $p$, and then $$h^\prime(p) = \left( \Re h \right)^\prime(p) + \iota \left( \Im h \right)^\prime(p);$$ that is, $$ \Re h^\prime(p) = \left( \Re h \right)^\prime(p) \ \mbox{ and } \ \Im h^\prime(p) = \left( \Im h \right)^\prime(p).$$ Finally, if $\lim_{x \to p} h(x) \neq 0$, then we can find a deleted neighborhood (or one-sided deleted neighborhood) of $p$ on which $h$ is non-zero. Am I right?

We will be using these facts in what follows.

From the situation given in the problem, we can conclude that $\Re f(x) \to 0$, $\Re g(x) \to 0$, $\Im f(x) \to 0$, $\Im g(x) \to 0$, $\Re f^\prime(x) \to \Re A$, $\Im f^\prime(x) \to \Im A$, $\Re g^\prime(x) \to \Re B$, and $\Im g^\prime(x) \to \Im B$ as $x \to 0$; moreover either $\Re B \neq 0$ or $\Im B \neq 0$. And, $\Re f$, $\Im f$, $\Re g$, $\Im g$, are all real differentiable functions on $(0, 1)$.

Then from Theorem 5.13, we can conclude that $$ \lim_{x \to 0} \frac{\Re f(x)}{x} = \lim_{x \to 0} \frac{ \Re f^\prime(x)}{1} = \Re A,$$ $$ \lim_{x \to 0} \frac{\Im f(x)}{x} = \lim_{x \to 0} \frac{ \Im f^\prime(x)}{1} = \Im A,$$ $$ \lim_{x \to 0} \frac{\Re g(x)}{x} = \lim_{x \to 0} \frac{ \Re g^\prime(x)}{1} = \Re B,$$ and $$ \lim_{x \to 0} \frac{\Im g(x)}{x} = \lim_{x \to 0} \frac{ \Im g^\prime(x)}{1} = \Im B;$$ thus $$ \lim_{x \to 0} \frac{ f(x) }{x } = \lim_{x \to 0} \frac{\Re f(x)}{x} + \iota \lim_{x \to 0} \frac{ \Im f(x)}{x} = \Re A + \iota \Im A = A, $$ and $$ \lim_{x \to 0} \frac{ g(x) }{x } = \lim_{x \to 0} \frac{\Re g(x)}{x} + \iota \lim_{x \to 0} \frac{ \Im g(x)}{x} = \Re B + \iota \Im B = B; $$ moreover since $B \neq 0$, therefore $$ \lim_{x \to 0} \frac{ x }{g(x) } = \frac{1}{B}, $$ and hence $$ \begin{align} \lim_{x \to 0} \frac{ f(x) }{ g(x)} &= \lim_{x \to 0} \left[ \left\{ \frac{f(x)}{x} - A \right\} \cdot \frac{x}{g(x)} + A \cdot \frac{x}{g(x)} \right] \\ &= \left\{ \lim_{x \to 0} \left( \frac{f(x)}{x} \right) - A \right\} \cdot \lim_{x \to 0} \left( \frac{x}{g(x)}\right) + A \cdot \lim_{x \to 0} \left( \frac{x}{g(x)} \right) \\ &= (A-A)\cdot \frac{1}{B} + A \cdot \frac{1}{B} \\ &= \frac{A}{B}. \end{align} $$

Is this proof correct and as intended by Rudin? If so, then is my reasoning correct? And, is my presentation rigorous enough as well?

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  • $\begingroup$ Your proof is very systematic and correct and exactly as expected by Rudin. I have seen some of your recent questions. Why do you doubt yourself when most of your own solutions are correct and reasonably good? $\endgroup$ – Paramanand Singh May 6 '17 at 10:19
  • $\begingroup$ Btw you should ask why the usual L'Hospital's Rule is not valid for complex functions of a real variable. $\endgroup$ – Paramanand Singh May 6 '17 at 10:22
  • $\begingroup$ @ParamanandSingh thanks for the encouraging comments. The reason why I find it to be a good idea checking with the Math SE community is one can err in one's thinking. For example, prior to the edit in this particular post, I had made a serious blunder in my origin post, which I managed to figure out and rectify on my own. But lest I do not manage to find out some such fallacy in my reasoning! $\endgroup$ – Saaqib Mahmood May 6 '17 at 18:30
  • $\begingroup$ @ParamanandSingh has this not got something to do with the fact that $\mathbb{R}$ is an ordered field, a property lacked by $\mathbb{C}$? I'm not sure if that is what the matter really is! $\endgroup$ – Saaqib Mahmood May 6 '17 at 18:34
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    $\begingroup$ You are right. There is no order relation in $\mathbb{C} $ due to which the Rolle's theorem is not applicable. Hence all those mean value theorems and L'Hospital's Rule are also not available. $\endgroup$ – Paramanand Singh May 7 '17 at 4:30

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