I was trying to come up with a proof of the well-known limit of products theorem. Suppose we have(adsbygoogle = window.adsbygoogle || []).push({});

$$\left|F_1 (t)-L_1\right|<\epsilon _1\Rightarrow \delta _1>|t-c|>0$$

and

$$\left|F_2 (t)-L_2\right|<\epsilon _2\Rightarrow \delta _2>|t-c|>0$$

$$\left|F_1 (t)-L_1\right|\left|F_2 (t)-L_2\right|<\epsilon _1\epsilon _2\Rightarrow min(\delta _1,\delta _2)>|t-c|>0$$

I could wave my hands and say that a well-behaved function approaches linearity in the neighborhood of ##c##, or something that sounds similarly impressive, but I am not satisfied with that. It seems my method of proof requires that any ##\delta < \delta _1## will also satisfy ##\left|F_1 (t)-L_1\right|<\epsilon _1##.

What justification, if any, have I for such an assertion? Again, thisseemsobvious for "well-behaved" functions, but I'm not sure which is the chicken and which is the egg.

I acknowledge that my proof seems a bit frumpy, even if it works; but the question about ##\delta < \delta _1## remains.

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# If d>|t-c|>0 satisfies e>|f(t)-L| does d1<d also?

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