Proving $\lim_{x\to c} g(x)$ exists given $\lim_{x\to c} f(x)$, $\lim_{x\to c} f(x)+g(x)$ do. Let $f,g$ be functions from $A$ to $\mathbb R$, and let $c$ be a cluster point of $A$. Show that if both $\lim_{x\to c} f $ and $\lim_{x\to c}(f+g)$ exist, then $\lim_{x\to c} g $  exists.
I need help with this proof, and here is my idea: Since both $\lim_{x\to c} f $ and $\lim_{x\to c}(f+g)$ exist, so $f(x)$ and $(f+g)(x)$ are bounded on a neighbourhood of c, and there exists a δ- neighbourhood $V_δ(c)$ of $c$ and a constant $M>0$ such that we have $|f(x)|\leq M$ and  $|(f+g)(x)|\leq M$. Then we get $|g(x)|\leq M$.
 A: HINT: $$|g(c) - g(y)| = |g(c) + f(c) - g(y) - f(y) + f(y) - f(c)|$$$$ \leq |g(c) + f(c) - g(y) - f(y)| + |f(y) - f(c)|$$
A: You can prove that if $\lim_{n\to\infty} a_n = a, \lim_{n\to\infty} b_n = b$ then $\lim_{n\to\infty} a_n + b_n = a+b$ exists.
Apply this to a fixed but arbitrary sequence $x_n$ with $x_n\to c$ and take $a_n = (f+g)(x_n)$ and $b_n = -f(x_n)$.
A: If $\lim_{x \to c} f(x) = a$ and $\lim_{x \to c} h(x) = b$ then $\lim_{x \to c} f(x) +h(x)= a+b$ and $\lim_{x \to c} -f(x) = -a$ (by the way, both exists).
So with $h(x)=f(x)+g(x)$ we get
$$\lim_{x \to c} g(x) = \lim_{x \to c} (g(x)+f(x))-f(x)= \lim_{x \to c} (g(x)+f(x))-\lim_{x \to c} f(x)$$
which exists.
A: Introduce two neigborhoods $V_1, V_2$ of $c$ such as:
$$
x\in V_1 \implies |f(x) - \lim_c f| < \epsilon
\\x\in V_2 \implies |f(x) + g(x) - \lim_c (f+g)| < \epsilon
$$
now for each $x\in V_1\cap V_2$,
$$
|g(x) - \lim_c (f+g) + \lim_c f|= 
|f(x) + g(x) - \lim_c (f+g) - [f(x) - \lim_c f]|\le 2\epsilon
$$
and as $V_1\cap V_2 $ is a neighborhood of $c$, you are done.
