from a previous question asked here Proving that an additive function $f$ is continuous if it is continuous at a single point someone stated that the key step is that $\lim_{x \to c}f(x)=\lim_{x \to a}f(x-a+c)$ however I don't understand how that is true shouldn't $\lim_{x \to c}f(x)=\lim_{x \to a}f(x+a-c)$ ? Thanks for you help!
Asked
Active
Viewed 45 times
0
-
First equation is correct and you can convince yourself by putting $t=x-a+c$ so that when $x\to a$ we have $t\to c$. – Paramanand Singh Oct 09 '17 at 17:55
-
why is the first equation correct though, what is the rationale to getting up to that point? – Skrrrrrtttt Oct 09 '17 at 17:58
-
Did you try the substitution $t=x-a+c$? It is obvious once you try this. – Paramanand Singh Oct 09 '17 at 17:59
-
if $f(x-a+c)$ approaches a then |(x-a+c)-a| by the definition of a limit which does not equal |x-c| – Skrrrrrtttt Oct 09 '17 at 18:22
-
$t=x-a+c$ approaches $c$ (and not $a$ as you think) when $x\to a$ and vice versa. Perhaps you are trying to think too deeply about trivial issues. – Paramanand Singh Oct 10 '17 at 04:36
1 Answers
0
I think what is confusing you is the re-use of the variable $x$.
In fact, this $x$ is a dummy variable.
$\displaystyle \lim\limits_{x\to c}f(x)$ or $\lim\limits_{y\to c}f(y)$ or $\lim\limits_{foo\to c}f(foo)$ are exactly the same thing, right ?
Now let's rewrite the correct statement using different dummy variables:
$\displaystyle \lim\limits_{x\to c}f(x)=\lim\limits_{y\to a}f(y-a+c)$
Intuititvely $x\to c$ means $x=c+\varepsilon$ where $\varepsilon$ is a very small quantity.
Yet $(y-a)$ is also such a very small quantity.
Thus in the substitution $x=c+\overbrace{(y-a)}^{\varepsilon}$ then $y\to a\iff x\to c$
The confusion arised because the dummy variable $y$ was changed back to $x$ leading to:
$\displaystyle \lim\limits_{x\to c}f(x)=\lim\limits_{x\to a}f(x-a+c)$
but you should really see it as a substitution like I explained before.
zwim
- 28,563
-
-
it just means that if $y$ get very close to $a$ then $(y-a)$ is very close to $0$ so $x$ is very close to $c$ – zwim Oct 09 '17 at 18:49
-
$\lim_{x\to c}f(x) = \lim_{h\to 0}f(x+h)$ can this concept also be used? to just replace h with y-a? – Skrrrrrtttt Oct 09 '17 at 18:52
-
This would be $\lim_{x\to c}f(x)=\lim_{h\to 0}f(c+h)$ but yes, just replace $h$ with $y-a$ and you get the result. – zwim Oct 09 '17 at 19:02