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An exercise I came across in my calculus text is as follows:

Prove that $f$ is continuous at $a$ if and only if $$\lim_{h\to0}f(a+h)=f(a)\tag{1}.$$

Now, I saw a proof of the Product Rule recently where the claim was that $g(x+h)\to g(x)$ as $h\to0$ since $g$ is differentiable and therefore continuous. Thus, I am wondering if it is fair to reformulate $(1)$ as $$ \lim_{h\to0}f(x+h)=f(x),\tag{2} $$ where $f$ is a differentiable and therefore continuous function. Would this not be accurate to say? My reading is that $(1)$ applies to all $a$ in the domain of $f$ and $x$ in $(2)$ simply indicates all values of $x$ in the domain of $f$ as well. Are $(1)$ and $(2)$ really the same thing? Does $(2)$ need any additional justification?

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  • $\begingroup$ The function $f$ may not be differentiable at $a$, but still be continuous there such as $f(x) = |x|$ when $x = 0$. How is continuity at a point defined in the calculus text? Just prove that the definition in the text can be reformulated as the limit you quoted. $\endgroup$ – Frank Hubeny Apr 30 '16 at 18:17
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    $\begingroup$ Somebody down-voted this. Doing that without explaining what is objectionable about the question is worthless and anti-social. $\qquad$ $\endgroup$ – Michael Hardy Apr 30 '16 at 18:18
  • $\begingroup$ @Michael Hardy. About downvoting (it's not me by I considered doing so, before I understood what the issue is) The problem is that the title (but also the beginning of the text) looks a little "simple-minded"... The title should use the word "vicinity" or say "for all x in a certain (a-k,a+k)" ! $\endgroup$ – Jean Marie Apr 30 '16 at 18:33
  • $\begingroup$ @JeanMarie : I believe you're mistaken. There's no need for this to hold for all $x$ in some open interval in order for this to make sense. In my answer I said "in the domain". There's no need for the domain to contain an interval for that to make sense. $\qquad$ $\endgroup$ – Michael Hardy Apr 30 '16 at 18:39
  • $\begingroup$ @Michael Hardy I agree $\endgroup$ – Jean Marie Apr 30 '16 at 18:41
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$$ \sum_{i=1}^5 i^2 = \underbrace{1^2+2^2+3^2 +4^2+5^2}_{\text{No “$i$'' or “$j$'' is seen here !}} = \sum_{j=1}^5 j^2. $$

In the expression above, $i$ and $j$ are $\text{“}$bound$\text{''}$ variables rather than $\text{“}$free$\text{''}$ variables. You can freely rename a bound variable without changing the meaning (as long as it doesn't conflict with another specification of the meaning of some variable).

\begin{align} & \text{For every value of $a$ in the domain } \lim_{h\to0} f(a+h) = f(a). \tag 1 \\[10pt] & \text{For every value of $x$ in the domain } \lim_{h\to0} f(x+h) = f(x). \tag 2 \end{align}

In the statements above, $a$ and $x$ are bound variables. The expressions $\text{“}$For every value of $a$ in the domain$\text{''}$ and $\text{“}$For every value of $x$ in the domain$\text{''}$ bind these variables just as $\text{“}\sum_{i=1}^5\text{''}$ and $\text{“}\sum_{j=1}^5\text{''}$ bind the variables $i$ and $j$. So $(1)$ and $(2)$ both say the same thing.

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  • $\begingroup$ Michael ... +1 ... Very well explained!! -Mark $\endgroup$ – Mark Viola Apr 30 '16 at 18:54
  • $\begingroup$ @Dr.MV : Thank you. $\qquad$ $\endgroup$ – Michael Hardy Apr 30 '16 at 19:00

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