How to Find the Limit Correctly? I am refreshing my Calculus memory, and bump into this example (need feedback):
$$f(x)= 
  \begin{cases}
   -x,    & \text{if } x < 0 \\
   x,        & \text{if } 0 \le x < 1 \\
   1 + x,    & \text{if } x \ge 1
  \end{cases}$$
I have to find the limits / state if it does not exist. Correct me if I am wrong:

*

*$\lim_{x \rightarrow 0} f(x)$ does not exist


*$\lim_{x \rightarrow 1} f(x)$  does not exist


*$f(1) = 2$


*$\lim_{x \rightarrow 1^{+}} f(x) = 2$
Are my answers correct?
This is the plot

 A: The limit of $f$ for $x \to 0$ exists and is $0$. Your other answers are correct.
A: As others have pointed out, the limit at $0$ exists. $f$ is equal to $|x|$ for $x < 1$. Since $|x|$ is continuous at $0$ so is this function.
Formally, $\forall \epsilon \in (0,1), |x|< \epsilon \implies |f|<\epsilon \implies \lim_{x\to 0} f(x) = 0$
A: This is (presumably) what you've plotted:
$$
y = 
\begin{cases}
 -x,    & \text{if } x < -1 \\
 x,        & \text{if } 0 \le x < 1 \\
 1 + x,    & \text{if } x \ge 1
\end{cases}
$$
And this is what the graph should have looked like.

Relying on tools is nice, but it's important to make sure that they are doing what you want them to do. For example, if I did this graph by hand, I might add a little hollow circle at $\{1,1\}$ to indicate that when $x$ is $1$, $y$ isn't $1$, it's $2$.  As the other answers noted, it's entirely possible to solve this problem without a plot.
A: For the first one, we have that
$$\lim_{x \rightarrow 0^+} f(x)=\lim_{x \rightarrow 0^+} x =0$$
$$\lim_{x \rightarrow 0^-} f(x)=\lim_{x \rightarrow 0^-} -x =0$$
therefore, since both one side limit are equal, we can conclude that
$$\lim_{x \rightarrow 0} f(x)=0$$
