Q: Is $f(x)$ continuous at $x=0$? For $f(x) = \frac{x^3+x}{x}$. Here's my work flow.

I've used this graph from Khan Academy, while it's indeterminate at first glance, it seems like we'd be able to find it's limit by factorising.


I also graphed it to see that there's any break in the graph, but can't seem to find a hole/break in the graph that makes is "not continuous", I've thus assumed that it IS continuous.
MY QUESTION:
What did I do wrong and can you help me identify/bridge my knowledge gap please?


Here's the original question, which doesn't even mention limit... :

If I cheated and used Wolfram alpha using limits it shows there's a break in limit, how did they get the graph like that?


A: For a function to be continuous at $x=a$ the condition is that $$\lim_{x \to a^-}f(x) = f(a) = \lim_{x \to a^+}f(x)$$ In the given case, we can easily see that, $$\lim_{x \to 0^-}f(x) = \lim_{x \to 0^-}\frac{x^3+x}{x} = \lim_{x \to 0^-}(x^2+1) = 1$$
And, $$\lim_{x \to 0^+}f(x) = \lim_{x \to 0^+}\frac{x^3+x}{x} = \lim_{x \to 0^+}(x^2+1) = 1$$ Now, if $f(0)$ were explicitly defined to be $1$, then the function would be continuous at $x=0$, as many others have pointed out.
In the absence of this definition, however, $$f(0) = \frac{0^3 + 0}{0}$$ which is undefined. The cancellation of $x$ in the numerator an denominator can only be done when solving for the limit, and not when finding out the functional value at that point.
Thus, $f(x)$ is not continuous at $x=0$

A: First of all, the condition for continuity of a function  $f(x) $ at a point $c$ is
$$\lim_{x\to c}f (x)=f(c) $$
Here, as you said the left hand side evaluates, by factoring, to $1$. However, the factorisation does work iff $x\neq 0$ (think why?). So, the function $f $ is not defined at $x=0$.
Clearly, as the right hand side is not defined, the condition is false, and therefore $f$ is discontinuous at $0$. The hole arises from the same.
Hope this helps. Ask anything if not clear :)

A: It really depends on what definitions you're using. For instance, for y=f(x) to be a function it needs to be defined for every x in its domain. In this case the domain hasn't been defined, so it's possible for it to be defined at f(0)=1, or it's possible the domain is R - {0}. Both of which are continuous (and these are the most likely ways to do things). It could also be defined as f(0)=10, which isn't continuous. However it's done, to be a function y=f(x) needs to be defined for every x in its domain.
The next thing that needs defined is the context of "continuity". Are you being asked whether the function is continuous over the Reals? Because it isn't unless f(0) is explicitly defined as 1. Or are you being asked if it's continuous over its domain? Because it is, in this case, if it's a function.
Generally, what you're actually being asked with questions like these is - is the domain the Real numbers? If it isn't then there are discontinuities over the Reals somewhere. Your job is to find and understand them.
In this case, we have a division by 0 that's created a removable discontinuity. So we can say that it is discontinuous over the Reals unless we define f(0)=1.
Please note that we can't define our way out of all our problems. Take 1/x, there's no value we can assign f(0) that will make it continuous at 0. Have a think about what makes it different.
