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- Homework Statement
- 10) c) Prove that ##\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} f(x^3)## (First see why the assertion is obvious; then provide a rigorous proof)

- Relevant Equations
- I suppose this problem isn't super hard. I find it tricky to explain the solution in words and then in symbols, in a way that is completely rigorous and doesn't assume any steps are so obvious that they can be skipped. My goal with the solution below is to try to actually prove the relationship in a way that is correct, precise, and doesn't leave absolutely anything implicit.

**c)**

**Why is the assertion ##\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} f(x^3)## obvious?**

First of all I don't think it is obvious but here is an explanation of why the limits are the same.

##\lim\limits_{x\to0} f(x^3)=l_2## means we are looking at points with ##x## close to zero and evaluating the function ##f(x^3)## at such points, checking if it stays close to some value ##l_2##.

Whereas in ##\lim\limits_{x \to 0}f(x)## we are evaluating ##f## at points ##x## close to zero, in ##\lim\limits_{x\to0} f(x^3)## we are evaluating ##f## at points ##x^3## with ##x## close to zero.

Keeping ##x## a distance ##\delta_1=\sqrt[3]{\delta}## from zero but then evaluating ##f## at points ##x^3## is the same as keeping ##x## a distance ##\delta## from zero, but then evaluating ##f## at points ##x##.

Therefore, if ##|f(x)-l|<\epsilon## when ##|x-0|<\delta## then ##|f(x^3)-l|<\epsilon## when ##|x^3-0|<\sqrt[3]{\delta}##, which we can write as ##\lim\limits_{y \to 0} f(y)=l##, for ##y=x^3##.

**Proof**

##\lim\limits_{x \to 0}f(x)=l## means that ##\forall \epsilon>0##, ##\exists \delta>0## such that ##|x-0|<\delta \implies |f(x)-l|<\epsilon##

##\lim\limits_{x \to 0} f(x^3)=l_2## means that ##\forall \epsilon>0##, ##\exists \delta_1>0## such that ##|x-0|<\delta_1 \implies |f(x^3)-l_2|<\epsilon##.

Let ##y=x^3 \implies x = \sqrt[3]{y}##

##|\sqrt[3]{y}|<\delta_1 \implies |y|<\delta_1^3=\delta_2 \implies |f(y)-l_2|<\epsilon##

##\implies \lim\limits_{y \to 0} f(y) = l_2##

##\implies l_2 = l_1##

Notice that ##\delta_1^3=\delta \implies \delta_1 = \sqrt[3]{\delta}##

If ##0<\delta<1## then ##\delta_1>\delta##, if ##0<1<\delta## then ##\delta_1<\delta##. This is interesting but is not important in the proof.

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