if $\lim_{x \to c}f(x)$ and $\lim_{x \to c }f(x)+g(x)$ exist prove that $\lim_{x \to c}g(x)$exist Please help me I stuck at this problem
if I prove by let $\lim_{x \to c}f(x)$ exist and 
but $\lim_{x \to c}g(x)$ does not exist 
$\lim_{x \to c}f(x)+g(x)$ can be exist 
we have 
$0<x-c<\delta \rightarrow |f(x)-L|<\frac{\epsilon }{2}$
but limit g(x) does not exist 
we will have
$0<x-c<\delta \rightarrow |g(x)|\geq \frac{\epsilon }{2}$
is it follow that by by this ?
$|f(x)-L|+|g(x)|\geq \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon $ that lead to contradict then limit g(x) must exist too
 A: Let $$\lim_{x \rightarrow c} f(x) = L_1$$
and $$\lim_{x \rightarrow c} \{f(x) + g(x)\} = L_2$$
Then $$\lim_{x \rightarrow c} g(x) = \lim_{x \rightarrow c} \{f(x)+g(x)\} - \lim_{x \rightarrow c} f(x)  = L_2-L_1$$
Note that it is possible that both $\lim_{x \rightarrow c} f(x)$ and $\lim_{x \rightarrow c} g(x)$ don't exist but $\lim_{x \rightarrow c} \{f(x)+g(x)\}$ exists.
Edit: You mentioned in the comments that you want to prove it ab initio, so I'm adding that proof too.( it is actually sufficient to prove sum theorem)
Let $\epsilon >0$ be given. As $\lim_{x \rightarrow c} f(x) = L_1$, there exists $\delta_1 >0$ such that $$0<|x-c|<\delta_1 \implies |f(x) - L_1| <\frac{\epsilon}{2}$$
Also as $\lim_{x \rightarrow c} \{f(x) + g(x)\} = L_2$, there exits $\delta_2 >0$ such that
$$0<|x-c|<\delta_2 \implies |f(x)+g(x) - L_2| <\frac{\epsilon}{2}$$
Let $\delta = \min\{\delta_1,\delta_2\}$ and $0<|x-c|<\delta$
Then $$|g(x)-(L_2-L_1)|\leq|f(x)+g(x)-L_2|+|f(x)-L_1| < \epsilon$$
A: Hint
We denote $\ell=\lim_{x\to c}f(x)$ and $\ell'=\lim_{x\to c}f(x)+g(x)$. You have to show that $\lim_{x\to c}g(x)=\ell'-\ell$. Since $g(x)=f(x)+g(x)-f(x)$ the result is almost clear. 
