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Category:Android (operating system) devicesQ:

If $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$, then $\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$

In the following theorem proved by Sierpinski, it is stated that:

If $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$, then $\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$.

I am curious about the method used to prove it, and why only use $x=0$ instead of any other $x$. It seems like the author simply assumed it. Is there any proof that does not use this assumption, or does this assumption hold for any $x$ not equal to $0$?

A:

Without loss of generality, assume $\lim_{x\rightarrow a} f(x) = L$ and $\lim_{x\rightarrow a} g(x) = M$.
Then $\lim_{x\rightarrow a} f(x) + g(x) = (L + M)$.
Since the conclusion is true when $f(x) = 0$, and since the limit exists, then it is true when $f(x) = q(x)g(x)$, and since the limit exists, and since the function $f$ is linear, the limit is true for any combination of limits of the form $f(x)g(x)$.
Then, we conclude the limit is true for $f(x) + g(x)$.
Since this holds for any such function $f$, the limit is true for \$f(x)
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