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Spivak, Ch 5 Limits, Problems 10c: Proving limit relationship
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[QUOTE="zenterix, post: 6578124, member: 695508"] Perhaps not obvious but also not not obvious. I am only using concepts from the current and previous chapter I am on in Spivak, so no bijections yet. Also not inverse functions, which is what I think might be needed to consider the case of replacing ##x^3## with ##\sin{x}##. However, using inverse sin, I think the proof is similar to the ##x^3## case: $$\lim\limits_{x \to 0} f(\sin{x})=l$$ means ##\forall \epsilon>0, \exists \delta>0: |x|<\delta \implies |f(\sin{x})-l|<\epsilon## Let ##y=\sin{x}##. Then ##x=\sin^{-1}{y}##, ##y \in [-1,1]##. $$(|x|<\delta \implies |\sin^{-1}{y}|<\delta \implies |y|<\sin{\delta}) \implies |f(y)-l|<\epsilon$$ $$\implies \lim\limits_{y \to 0} f(y) = l$$ The gist of these proofs is: Assuming ##\lim\limits_{x \to 0} f(\sin{x})## exists, then keeping ##x## within ##\delta## of ##0## means ##f(\sin{x})## is within ##\epsilon## of ##l##. But if ##x## is actually a function ##x=\sin^{-1}{y}## of ##y##, then we can determine a ##\delta_1## such that if ##y## is within ##\delta_1=\sin{\delta}## of ##0##, then ##x## is within ##\delta## of ##0##, and so ##f(\sin{x})=f(y)## is within ##\epsilon## of ##l##. In this proof, I indeed assumed that both limits exist, in particular I assumed ##\lim\limits_{x \to 0} f(\sin{x})## exists. Of course if the second limit in ##\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} f(\sin{x})## doesn't exist then the statement is false, so we can't prove that it is true. If we had instead replaced ##x^3## with ##\frac{1}{x}## we would have had to assume that ##\lim\limits_{x \to 0} f(\frac{1}{x})## exists. An interesting question is, for what functions does it actually exist? Clearly not all of them: ##f(x)=x \implies f(\frac{1}{x})=\frac{1}{x}##, which does not have a defined limit at ##x=0##. As an example of a function where the limit does exist, consider ##f(x)=\frac{1}{1+x}##. ##f(\frac{1}{x})=\frac{1}{\frac{1}{x}+1}=\frac{x}{1+x}## and ##\lim\limits_{x \to 0} f(\frac{1}{x}) = 0##. If we assume we are dealing with one of the ##f(\frac{1}{x})## for which the limit does exist, we have: $$\lim\limits_{x \to 0} f(\frac{1}{x})=l_2$$ means ##\forall \epsilon > 0 \exists \delta>0 : |x|<\delta \implies |f(\frac{1}{x})-l_2|<\epsilon## Let ##y=\frac{1}{x}##. Then ##x=\frac{1}{y}##. $$(|x|<\delta \implies |\frac{1}{y}|<\delta \implies |y|>\frac{1}{\delta}) \implies |f(y)-l_2|<\epsilon$$ $$\implies \lim\limits_{y \to \infty}f(y) = l_2$$ However, I am not sure how to reconcile this with $$\lim\limits_{x \to 0} f(x)=l$$ Such that I can conclude that ##l=l_2##. It is clear intuitively, but I mean notation-wise and in terms of the definitions, how do I reconcile the two? [/QUOTE]
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Spivak, Ch 5 Limits, Problems 10c: Proving limit relationship
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