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. 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
 A: 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 $\lim_{x\to 0}g(x)=\infty$
A: The key is that you can only say two limits are equal if they exist, and the two limits on the RHS cannot be guaranteed to exist with the given information. You are right in suggesting that the answer can involve infinity, but there are finite examples that are badly behaved as well.
Consider the function $f(x)$ where $f(x)=1$ for $x$ rational and $f(x)=0$ for $x$ irrational. It is a well defined function, but has no limit anywhere. If we let $g(x)=f(x)$, then $f(x)-g(x)=0$ everywhere so the limit is zero for every $\alpha$, but $f(x)$ and $g(x)$ have no limit no matter what value of $\alpha$ is chosen.
A: What about $\alpha =0$ and
$$f(x)=1+\sin\left(\frac{1}{x}\right) \text{ and } g(x)= \sin\left(\frac{1}{x}\right)$$
as a counterexample? If you want $f,g$ to be defined at $0$ just set $f(0)=g(0)=0$.
If you suppose either $f$ or $g$ to have a limit at $\alpha$, then the result becomes true.
A: Take $f=g$ to be any function that does not have a limit at $\alpha$, for example $\tan x$ at $\alpha=\frac\pi2$.
A: Start from a simple case in which this is true:
$$ f=x, \quad g=0 $$ when $ \alpha \to 0 $.
We do have the limit $$ \lim_{\alpha \to 0} f(x)-g(x)=0 $$
For any situation similar to this, I can simply plus a so-called "pathological" term to both $f$ and $g$ to make their limits at $0$ do not exist
$$ f=x+\sin \frac{1}{x}, \quad g=\sin \frac{1}{x} $$
or
$$ f=x+e^{\frac{1}{x}}\sin \frac{1}{x^2}, \quad g=e^{\frac{1}{x}}\sin \frac{1}{x^2} $$
Thus our counterexamples do not have to be infinity cases.
Hope this would help.
