[Spivak Calculus, Ch. 5 P. 9] Showing equality of two limits

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SUMMARY

The discussion centers on proving the equality of two limits: ##\lim_{x \rightarrow a} f(x) = \lim_{h \rightarrow 0} f(a+h)##. The proof relies on the definitions of limits, where for any ##\epsilon > 0##, there exist ##\delta_1## and ##\delta_2## that satisfy the conditions for both limits. The user suggests a proof by contradiction, assuming ##l \neq m##, but seeks clarity on the contradiction needed to establish the equality of the limits. The goal is to demonstrate that if one limit exists, the other must also exist and be equal.

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Homework Statement



Prove that ##\lim_{x \rightarrow a} f(x) = \lim_{h \rightarrow 0} f(a+h)##.

Homework Equations



By definition, if ##\lim_{x \rightarrow a} f(x) = l## then for every ##\epsilon > 0## there exists some ##\delta_1## such that for all x, if ##0<|x-a|<\delta_1## then ##|f(x)-l|<\epsilon##.

Similarly, if ##\lim_{h \rightarrow 0} f(a+h) = m## then for every ##\epsilon > 0## there exists some ##\delta_2## such that for all h, if ##0<|h-0|<\delta_2## then ##|f(a+h)-m|<\epsilon##.

The Attempt at a Solution



I'm really not sure how to go from here. I think maybe I need to perform a proof by contradiction by assuming that ##l \neq m##, but I don't know what kind of contradiction I'm looking for.
 
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I think you need to prove that if the LHS has some limit l, then the RHS also has that same limit, and vice versa.
 
Edit: actually, not quite sure what he's asking for here.
 

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