Theorem
Let f:R2→R and (a,b)∈R2. Let fb:R→R be defined by fb(x)=(x,b) for all x∈R. Then, lim(x,y)→(a,b)f(x,y)=f(a,b)⟺(lim(x,y)→(a,b)g(x,y)=0)∧(limx→afb(x)=fb(a))
where g:R2→R defined by g(x,y)=f(x,y)−f(x,b).
My Proof
Necessity Part
Let us first prove that, lim(x,y)→(a,b)f(x,y)=f(a,b)⟹(lim(x,y)→(a,b)g(x,y)=0)∧(limx→afb(x)=fb(a))
Let ((xn,yn))n≥1 be any sequence converging to (a,b). Then since lim(x,y)→(a,b)f(x,y)=f(a,b), by definition we have, (∀ε>0)(∃n1(ε)∈N)[n≥n1(ε)⟹|f(xn,yn)−f(a,b)|<ε]Furthermore since fb is continuous at x=a (I am skipping the proof of it for now) we have, (∀ε>0)(∃n2(ε)∈N)[n≥n2(ε)⟹|f(xn,b)−f(a,b)|<ε]What remains is to prove that, lim(x,y)→(a,b)g(x,y)=0. For this observe that from (1) and (2) we can write, (∀ε2>0)(∃n1(ε2)∈N)[n≥n1(ε2)⟹|f(xn,yn)−f(a,b)|<ε2](∀ε2>0)(∃n2(ε2)∈N)[n≥n2(ε2)⟹|f(xn,b)−f(a,b)|<ε2]Now taking N(ε2)=max(n1(ε2),n2(ε2)) we get, (∀ε2>0)(∃N(ε2)∈N)[n≥N(ε2)⟹|f(xn,yn)−f(a,b)|<ε2](∀ε2>0)(∃N(ε2)∈N)[n≥N(ε2)⟹|f(xn,b)−f(a,b)|<ε2]From (3) and (4) and using Triangle Inequality we get, (∀ε>0)(∃N(ε2)∈N)[n≥N(ε2)⟹|f(xn,yn)−f(xn,b)|<ε]In other words, (∀ε>0)(∃N(ε2)∈N)[n≥N(ε2)⟹|g(xn,yn)−g(a,b)|<ε]and this is what we wanted to prove.
Sufficiency Part
Now we try to prove that, lim(x,y)→(a,b)f(x,y)=f(a,b)⟸(lim(x,y)→(a,b)g(x,y)=0)∧(limx→afb(x)=fb(a))
Let ((xn,yn))n≥1 be any sequence converging to (a,b). Observe that since g is continuous at (a,b) we have, (∀ε>0)(∃n1(ε)∈N)[n≥n1(ε)⟹|g(xn,yn)−g(a,b)|<ε]In other words, (∀ε>0)(∃n1(ε)∈N)[n≥n1(ε)⟹|f(xn,yn)−f(xn,b)|<ε]Also, since fb is continuous at a, we have, (∀ε>0)(∃n2(ε)∈N)[n≥n2(ε)⟹|f(xn,b)−f(a,b)|<ε]Now observe that from (5) and (6) we can conclude respectively that, (∀ε2>0)(∃n1(ε2)∈N)[n≥n1(ε2)⟹|f(xn,yn)−f(xn,b)|<ε2](∀ε2>0)(∃n2(ε2)∈N)[n≥n2(ε2)⟹|f(xn,b)−f(a,b)|<ε2]
Now taking N(ε2)=max(n1(ε2),n2(ε2)) from the above two statements (using Triangle Inequality) we can conclude that, (∀ε>0)(∃N(ε2)∈N)[n≥N(ε2)⟹|f(xn,yn)−f(a,b)|<ε]Since ((xn,yn))n≥1 was arbitrary, we are done.
Answer
Yes, your proof is entirely correct, allthough it is hard to read because of your quantor notation. But mathematically it is completely valid. Good Job!
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