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I am looking at the proof that differentiability implies continuity:

$0 = 0 \cdot f' ( z_0) = \lim_{z \to z_0} (z - z_0) \cdot \lim_{z \to z_0}\dfrac {f(z) - f(z_0)}{z - z_0} = \lim_{z \to z_0} (z - z_0) \dfrac {f(z) - f(z_0)}{z - z_0} = \lim_{z \to z_0} [f(z)- f(z_0)]$

The proof stops here, after proving that $\lim_{z \to z_0} [f(z)- f(z_0)] = 0$. However, the definition of continuity that is given in my book is $\lim_{z \to z_0} f(z) = f(z_0)$. We could easily arrive that this is we know that $\lim_{z \to z_0} f(z)$ exists, because then we can split up the limits. However, I am not sure how to prove that $\lim_{z \to z_0} f(z)$ exists.

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  • $\begingroup$ @dxiv I can see that if $f(z) - f(z_0)$ doesn't approach $0$, then the limit of the difference quotient is infinity. However, I have failed to prove this using $\epsilon -\delta$. Anyway, I am cofortable with this proof, except for the very last line. $\endgroup$ – Ovi Feb 22 '17 at 5:39
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Keep in mind that $f(z_0)$ is a constant and can be removed from the limit. So

$$\begin{array}{cc} &\lim_{z \to z_0}[f(z)-f(z_0)]+f(z_0)=0+f(z_0)\\ \Rightarrow & \lim_{z \to z_0}[f(z)-f(z_0)+f(z_0)]=f(z_0)\\ \Rightarrow & \lim_{z \to z_0}f(z)=f(z_0) \end{array}$$ as desired.

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  • $\begingroup$ ovi's right you need to use an epsilon delta proof to truly prove this analytically. $\endgroup$ – Sentinel135 Feb 22 '17 at 5:44
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    $\begingroup$ Well I have proved the propreties of limits using $\epsilon - \delta$, so I think I have earned the right to use them :) $\endgroup$ – Ovi Feb 22 '17 at 5:46
  • $\begingroup$ @ovi: +1 for your comment about "earned the right". $\endgroup$ – Paramanand Singh Feb 22 '17 at 8:34
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Recall from the definition of limit, to prove $\lim_{z\to z_0} f(z)=f(z_0)$ you are tasked with showing for all $\epsilon >0$ there is a $\delta>0$ such that$$ |f(z)-f(z_0)| < \epsilon$$ whenever $|z-z_0|<\delta.$

From the same definition, to prove $\lim_{z\to z_0}(f(z)-f(z_0)) = 0$ you need to show for all $\epsilon >0$ there is a $\delta>0$ such that $$ |f(z)-f(z_0)-0| < \epsilon$$ whenever $|z-z_0|<\delta.$

Same thing. The two statements are equivalent.

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The computation that you quoted shows that $\lim\limits_{z\to z_0}[f(z)-f(z_0)]$ exists, and moreover it's equal to $0$. Also, $\lim\limits_{z\to z_0}f(z_0)$ exists as the limit of a constant function, and it's obviously equal to $f(z_0)$. Therefore, by properties of limits, the limit of the sum of these two functions exists as well: $\lim\limits_{z\to z_0}\left\{[f(z)-f(z_0)]+f(z_0)\right\}$ exists. And since $[f(z)-f(z_0)]+f(z_0)=f(z)$, we can deduce that $$\lim\limits_{z\to z_0}f(z)=\lim\limits_{z\to z_0}\left\{[f(z)-f(z_0)]+f(z_0)\right\}=0+f(z_0)=f(z_0).$$

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If $\lim_{z\to z_0}(f(z)-f(z_0))=0$, then given $\epsilon>0$ there exists a $\delta>0$ such that

$$|f(z)-f(z_0)|<\epsilon$$

whenever $|z-z_0|<\delta$.

But this is the definition of the limit $\lim_{z\to z_0} f(z)=f(z_0)$.

That is to say, the two statements, $\lim_{z\to z_0}(f(z)-f(z_0))=0$ and $\lim_{z\to z_0}f(z)=f(z_0)$, are equivalent.

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The proof is correct, and not circular.

$0 = 0 \cdot f' ( 0)$

$f'(0)$ exists because $f$ is differentiable.

$= \lim_{z \to z_0} (z - z_0) \cdot \lim_{z \to z_0}\dfrac {f(z) - f(z_0)}{z - z_0}$

  • $0=\lim_{z \to z_0} (z - z_0)$ because the identity function is continuous;

  • $f'(0) = \lim_{z \to z_0}\dfrac {f(z) - f(z_0)}{z - z_0}$ by definition of the derivative.

$ = \lim_{z \to z_0} (z - z_0) \dfrac {f(z) - f(z_0)}{z - z_0}$

Since the limits exist for both factors, the limit of their product exists and is the product of the two limits.

$= \lim_{z \to z_0} [f(z)- f(z_0)]$

This follows by algebraic simplification $\require{cancel}\lim_{z \to z_0} \cancel{(z - z_0)} \dfrac {f(z) - f(z_0)}{\cancel{z - z_0}} = \lim_{z \to z_0} \big(f(z)- f(z_0)\big)$.

Rolling all the equalities back to the beginning, the last line proves that:

$$0 = \lim_{z \to z_0} \big(f(z)- f(z_0)\big)$$

Therefore the function $g(z)=f(z)-f(z_0)$ is continuous at $z_0\,$, directly by definition of continuity.

Since the sum of continuous functions is continuous, and since the constant function $h(z)=f(z_0)$ is continuous, it follows that the sum $g(z)+h(z)=f(z)$ is continuous at $z_0\,$.

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This exactly does not address your concerns, but it is too long for a comment.


The way your proof proceeds is correct but very unusual (at least for me). All the stuff in that chain of equalities should have been reversed. To prove that $f(z)$ is continuous at $z_{0}$ you need to calculate its limit at $z_{0}$ and show that the limit is $f(z_{0})$. And it is better to proceed in forward fashion. \begin{align} \lim_{z \to z_{0}}f(z) &= \lim_{z \to z_{0}}\frac{f(z) - f(z_{0})}{z - z_{0}}\cdot (z - z_{0}) + f(z_{0})\notag\\ &= f'(z_{0})\cdot 0 + f(z_{0})\notag\\ &= f(z_{0})\notag \end{align} The approach of showing $\lim_{z \to z_{0}}(f(z) - f(z_{0})) = 0$ to deduce continuity of $f$ also seems a little contrived. To give a simple analogy, consider that $3x - 6 = 0$. Then I don't like to first prove $x - 2 = 0$ and then deduce that $x = 2$ when it is very easily possible to show that $x = 6/3 = 2$.

Whenever you are dealing with a limit problem, always always give it a shot via algebra of limits (which are far more powerful than most students think they are) and then only try other tools/techniques.

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  • $\begingroup$ Thank you for the insights, your proof is indeed much cleaner. At first I noticed that if $f(z)$ does not approach $f(z_0)$, then $\lim_{z \to z_0} \dfrac {f(z)-f(z_0)}{z-z_0}$ would be infinity. I thought there would be an easy way to prove this with $\epsilon-\delta$, but was unsuccesfull. I will follow your advice of using the algebra of limits. $\endgroup$ – Ovi Feb 22 '17 at 17:09

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