limit proof

Jan 2010
150
29
Mexico City
One way of seeing it:

Absolute value is a continuous function so \(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=\Big|\lim_{x \to 0} f(x) - \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\Big) \Big| =|0-0|=0\)

Where it was used the fact that since when\(\displaystyle x\to 0 \) then \(\displaystyle \frac{x}{2^n}\to 0 \) as well, we have
\(\displaystyle \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{\frac{x}{2^n} \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{y \to 0} f(y)=0 \)


Finally

\(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=0=\left|\lim_{x \to 0} f(x)\right |=\lim_{x \to 0} |f(x)|\)
 
Jan 2010
150
29
Mexico City
One way of seeing it:

Absolute value is a continuous function so \(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=\Big|\lim_{x \to 0} f(x) - \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\Big) \Big| =|0-0|=0\)

Where it was used the fact that since when\(\displaystyle x\to 0 \) then \(\displaystyle \frac{x}{2^n}\to 0 \) as well, we have
\(\displaystyle \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{\frac{x}{2^n} \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{y \to 0} f(y)=0 \)


Finally

\(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=0=\left|\lim_{x \to 0} f(x)\right |=\lim_{x \to 0} |f(x)|\)

Hmmm i totally missread the question, i am sorry let me check it again
 
Jan 2010
150
29
Mexico City
Certainly i dont understand how that step is done,

in fact it would seem to me that \(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)|\) doesnt make any sense since the left side does not depend on x.

Returning to the proof, earlier we had that

\(\displaystyle \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|<2\epsilon x\) for all n if \(\displaystyle 0< x <\alpha\)

Taking \(\displaystyle n\to\infty\) and using the continuity of the abs value we get that since if \(\displaystyle n\to \infty\) then \(\displaystyle y=\frac{x}{2^n}\to 0\),

in our case :
if \(\displaystyle n\to \infty \) then \(\displaystyle f(\frac{x}{2^n})=f(y)\to 0\).

And we end up having \(\displaystyle |f(x)|\leq 2\epsilon x \) from where the proof follows quite easily.
 
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May 2009
959
362
Certainly i dont understand how that step is done,

in fact it would seem to me that \(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)|\) doesnt make any sense since the left side does not depend on x.

Returning to the proof, earlier we had that

\(\displaystyle \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|<2\epsilon x\) for all n if \(\displaystyle 0< x <\alpha\)

Taking \(\displaystyle n\to\infty\) and using the continuity of the abs value we get that since if \(\displaystyle n\to \infty\) then \(\displaystyle y=\frac{x}{2^n}\to 0\),

in our case :
if \(\displaystyle n\to \infty \) then \(\displaystyle f(\frac{x}{2^n})=f(y)\to 0\).

And we end up having \(\displaystyle |f(x)|\leq 2\epsilon x \) from where the proof follows quite easily.
EDIT : So you think it's just a typo and what they're really doing is taking the limit as n approaches infinity?
 
Jan 2010
150
29
Mexico City
Either a typo or some argument (which is not clear, at least for me) equivalent to the n-limit one. So far i dont see any flaws on this one.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
In the following proof, why does the fact that \(\displaystyle \lim_{x \to 0} f(x) =0 \) imply that \(\displaystyle \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| \) ?

http://www.math.purdue.edu/pow/spring2005/pdf/solution9.pdf
It certainly looks to me like this was supposed to be
\(\displaystyle \lim_{n\to \infty}\left|f(x)- f(x/2^n)\right|= |f(x)|\)
Because the right hand side depends on x, not n. But, since that is \(\displaystyle \left|f(x)- \lim_{n\to\infty}f(x/2^n)\right|\) it looks obvious.
 
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