# limit proof

• May 22nd 2010, 11:20 AM
Random Variable
limit proof
In the following proof, why does the fact that $\lim_{x \to 0} f(x) =0$ imply that $\lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)|$ ?

http://www.math.purdue.edu/pow/sprin.../solution9.pdf
• May 22nd 2010, 12:06 PM
mabruka
One way of seeing it:

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

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

Finally

$\lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=0=\left|\lim_{x \to 0} f(x)\right |=\lim_{x \to 0} |f(x)|$
• May 22nd 2010, 12:08 PM
mabruka
Quote:

Originally Posted by mabruka
One way of seeing it:

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

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

Finally

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

Hmmm i totally missread the question, i am sorry let me check it again
• May 22nd 2010, 12:25 PM
mabruka
Certainly i dont understand how that step is done,

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

Returning to the proof, earlier we had that

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

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

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

And we end up having $|f(x)|\leq 2\epsilon x$ from where the proof follows quite easily.
• May 22nd 2010, 12:30 PM
Random Variable
Quote:

Originally Posted by mabruka
Certainly i dont understand how that step is done,

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

Returning to the proof, earlier we had that

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

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

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

And we end up having $|f(x)|\leq 2\epsilon x$ 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?
• May 22nd 2010, 01:08 PM
mabruka
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.
• May 23rd 2010, 03:39 AM
HallsofIvy
Quote:

Originally Posted by Random Variable
In the following proof, why does the fact that $\lim_{x \to 0} f(x) =0$ imply that $\lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)|$ ?

http://www.math.purdue.edu/pow/sprin.../solution9.pdf

It certainly looks to me like this was supposed to be
$\lim_{n\to \infty}\left|f(x)- f(x/2^n)\right|= |f(x)|$
Because the right hand side depends on x, not n. But, since that is $\left|f(x)- \lim_{n\to\infty}f(x/2^n)\right|$ it looks obvious.