Intuition behind limits I understand the concept of a limit, but here is my problem. Say I want to find the limit of some function $f(x)$ as $x\to0$. What checks/list should I go through to make sure the limit will exist? Is there an intuitive way to check if the limit will exist without having to check both left and right-hand limits? Based on my understanding, I would always have to check $x\to 0^-$ and $x\to 0^+$. However, with some limits, this is not necessary and we can just "plug-in", say for this example, $0$. I noticed in cases where we can just plug-in, $f(x)$ would have been in the form "$0/0$", requiring me to do some algebra to find the limit. In cases where, but is not restricted to, the limit as $x\to 0$ does not exist, $f(x)$ is in the form "$n/0$" where $n$ is some number other than $0$.
I hope that makes sense.
 A: It's hard (if not impossible) to capture "all cases" in a few simple guidelines, but based on your question I'll limit (pun not intended!) this answer to the most common things you'll encounter when taking a limit $x \to a$, where $a$ can be $0$ but not necessarily, of a function in the form of a fraction $\tfrac{f(x)}{g(x)}$. So you're looking at a limit of the form:
$$\lim_{x \to a} \frac{f(x)}{g(x)}$$
When you would simply plug in $x=a$, there are four cases regarding $0$ that can arise:


*

*$\frac{\mbox{non-zero}}{\mbox{non-zero}}$: there is no problem...!

*$\frac{\mbox{zero}}{\mbox{non-zero}}$: again, there is no problem: the fraction is simply $0$.

*$\frac{\mbox{non-zero}}{\mbox{zero}}$: in absolute value, the fraction will tend to $\infty$; check the right- and left-handed limits to see whether these are $\pm \infty$ (and they can have the same or a different sign).

*$\frac{\mbox{zero}}{\mbox{zero}}$: anything can happen here, further investigation is necessary.
In this last case, you'll typically try to eliminate the indeterminate form "$\tfrac{0}{0}$" by:


*

*factoring and canceling $x-a$ when possible, e.g. when $f$ and $g$ are polynomials;

*multiplying numerator and denominator with a suitable conjugate expression to get rid of (square or other) roots; again trying to cancel a factor of the form $x-a$ afterwards;

*using standard limits such as $\tfrac{\sin x}{x} \to 1$ as $x \to 0$;

*using techniques such as l'Hôpital's rule or (pieces of) Taylor series.


Note that this is a simplified overview (incomplete and not rigorous) but it probably covers most of the simple examples you encounter when you're new to calculating limits. Questions? Shoot!

Examples
I'll add an example for each of the four categories.


*

*Consider the following limit:
$$\lim_{x \to 0} \frac{\sqrt{3x+4}}{x^2-3}$$
Plugging in shows there's no problem in numerator nor denominator; the function is continuous in $x=0$ and you simply find the limit as:
$$\lim_{x \to 0} \frac{\sqrt{3x+4}}{x^2-3}=\frac{\sqrt{3\cdot 0+4}}{0^2-3}=\frac{\sqrt{4}}{-3} = -\frac{2}{3}$$

*Next up, one where only the numerator becomes $0$:
$$\lim_{x \to -2} \frac{x^2-4}{\sqrt{3-x}}=\frac{(-2)^2-4}{\sqrt{3-(-2)}}=\frac{0}{\sqrt{5}}  = 0$$

*Only the denominator tends to $0$:
$$\lim_{x \to 1} \frac{2x}{x-1}$$
Now notice that near $x=1$, the numerator is positive and the denominator is positive for $x>1$ but negative for $x<1$; hence:
$$\lim_{x \to 1^+} \frac{2x}{x-1} = +\infty \quad\mbox{and}\quad \lim_{x \to 1^-} \frac{2x}{x-1} = -\infty \implies \lim_{x \to 1} \frac{2x}{x-1} \;\mbox{"does not exist"}$$
Note that with a square in the denominator, we would have:
$$\lim_{x \to 1^+} \frac{2x}{\left( x-1 \right)^2} = +\infty \quad\mbox{and}\quad \lim_{x \to 1^-} \frac{2x}{\left( x-1 \right)^2} = +\infty \implies \lim_{x \to 1} \frac{2x}{\left( x-1 \right)^2}  = +\infty$$

*Plugging in gives the indeterminate form "$\tfrac{0}{0}$":
$$\lim_{x \to 2} \frac{x^2-4}{x^2-2x}$$
Factoring, canceling and trying again:
$$\lim_{x \to 2} \frac{x^2-4}{x^2-2x} = 
\lim_{x \to 2} \frac{\color{red}{(x-2)}(x+2)}{x\color{red}{(x-2)}}= 
\lim_{x \to 2} \frac{x+2}{x} = \frac{2+2}{2}=2$$

