# limit proof

#### mabruka

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

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

#### mabruka

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

#### mabruka

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.

Random Variable

#### Random Variable

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?

#### 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.

#### HallsofIvy

MHF Helper
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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