A few questions about limits and also continuity I have a test tomorrow and I've been studying for most of the week (including last) and I just want some clarification on a few things. If you guys can help that would be fantastic. 
So looking at this problem,
Limit : $x \rightarrow 0$
$$\frac{\frac{3}{x+3} - 1}{x} $$
If I were to do it by simply plugging it in, I would get $3/3 - 1$ which is $0$ and the denominator would obviously also be $0$ so the final answer would be $0/0$.
Now if I didn't want this answer, couldn't I just use $.001$ to find the limit instead?
I believe the answer when I do it this way is $-.33$
Can I just do this for all limits? Is this even the correct way to approach this problem?
Lastly, can someone clarify when a limit is continuous at $x = c$? 
I believe that there are 3 parameters and I can only recall two currently. 
 A: The final answer wouldn't be $\frac{0}{0}$, since that's undefined. You'd have to algebraically manipulate it in order to avoid the $\frac{0}{0}$ scenario. In your example above, simplifying the numerator, allows you to cancel with the denominator. $\dfrac{\frac{3}{x+3}-1}{x}=\dfrac{\frac{3}{x+3}-\frac{x+3}{x+3}}{x}=\dfrac{-\frac{x}{x+3}}{x}=-\dfrac{1}{x+3}$. After simplifying the expression, now evaluating this as $x\to 0$ is no problem.
You can't just substitute $-0.33$, since that's not close enough to $0$, but regardless of what value you choose that you consider close to $0$, I can always choose a number that's closer to $0$, so substituting values close to the limit numerically won't work.

A function $f(x)$ is said to be continuous at $x=c$ if:


*

*$\lim_{x\to c^-}f(x)=\lim_{x\to c^+} f(x)$ (This means that $\lim_{x\to c} f(x)$ exists.) 

*$\lim_{x\to c} f(x)=f(c)$ 
A: $$ \lim_{x\to 0} \frac{\frac{3}{x+3}-1}{x} = \lim_{x\to 0} \frac{\frac{3-x-3}{x+3}}{x}= \lim_{x\to 0} \frac{-\frac{x}{x+3}}{x} $$
$$= \lim_{x\to 0} -\frac{x}{x(x+3)} = \lim_{x\to 0} -\frac{1}{x+3} = -\frac{1}{0+3} =-\frac{1}{3}$$
