TITLE: True or False? If the limit of $[f(x)-g(x)]$ as x approaches $a = 0$, then the limit of $f(x), g(x)$, as $x$ approaches $a$, are equal.
QUESTION [8 upvotes]: TRUE OF FALSE?
$$\lim_{x\to\alpha}[f(x)-g(x)]=0\implies\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha} g(x)$$
I'm in senior high school and just recently took a test with this as one of the questions. I initially answered true — I mean, it obviously looks too true. Simple distribution and manipulation of terms would easily show $\lim_{x\to\alpha}f(x)=\lim_{x\to\alpha}g(x)$ — but just before submitting, I changed it to false. I got a perfect score, but I do feel uneasy not knowing why my teacher remarked this as false.
Just for context, maybe the counterexample to make this statement false might involve infinities.  When the limit is $+\infty$ or $-\infty$, we simply write is as "does not exist."
Also we only tackle real number. And our limit definition specifies that $\lim_{x\to\alpha}f(x)$ only exists when $\lim_{x\to\alpha^{+}}f(x)=\lim_{x\to\alpha^{-}}f(x)$.
Qn: The actual question looked like this Here. Link

REPLY [4 votes]: If atleast one of $\lim f(x)$ or $\lim g(x)$ exists on $\mathbb R$ i.e. exists finitely then the given statement is true.  
WLOG let $\lim_{x\to a}g(x)=l$, where $l\in \mathbb R$
Given that: $\lim_{x\to a}(f(x)-g(x))=0$ 
So $f(x)=(f(x)-g(x))+g(x)\implies \lim_{x\to a}f(x)=0+l=l=\lim_{x\to 0}g(x)$ 
However, if you consider $f(x)=[x]$, where $[.]$ is floor function and $g(x)=-\{x\}$, where $\{.\}$ is fractional part function. Indeed we have $\lim _{x\to 0} (f(x)-g(x))=0$ but neither $\lim_{x\to 0}f(x)$ exists nor does $\lim_{x\to 0} g(x)$. 
Note: It is to be noted that even if either of the limits $\lim_{x\to a} f(x)$ or $\lim_{x\to a} g(x)$ exists on extended reals that is on the set $\mathbb R\cup \{-\infty, \infty\}$, still if $\lim_{x\to a}(f(x)-g(x))=0$
then $f(x)$ and $g(x)$ have the same limiting behaviour as $x\to a$. 
For example: Consider $f(x)=x+\frac 1{|x|}, g(x)=\frac 1{|x|}$, then $f(x)-g(x)=x\to 0$ as $x\to 0$. Also, $\lim_{x\to 0}f(x)=\infty$ and $\lim_{x\to 0}g(x)=\infty$