Defined, and Limit Existence True or False Explanation

Question

I have the following true and false statement:

If $f(a)$=L, then the $\lim_\limits{x\to a}f(x)=L$.

I drew the following graph as a counterexample: The equations for the piecewise come from the following piecewise: \begin{equation}f(x)=\begin{cases}\frac{x^3-3x^2}{(x-3)}, x\ne 3\\ 5, x=3 \end{cases}\end{equation}

My Question

My question is if this is enough to write a general statement about why this statement is false, and would this be ok enough to cover the broadness of the types of solutions?

My Answer to the True and False Statement:

If the function were to have a removable discontinuity, and the limit existed and was not equal to the value where the function was defined thus this statement would be false. For continuous functions this is true.

Answer 1

Nitpick. The graph would be clearer if you had made the blue point red, and if you had put a hole in the graph of the rational function.

Your counter example is fine but it doesn't need to be be so hard.

It's enough to say; $\lim_{x\to a} f(x) = f(a)$ if and only if $f$ is continuous at $x = a$ and the result will not be true for any $f$ that is not continuous at $a$. For example let $\begin{equation}f(x)=\begin{cases} 0 & x\ne 1\newline 2 & x=1\end{cases}\end{equation}$ where $\lim_{x\to 1} f(x) = 0$ but $f(1) = 2$

Answer 2

The statement is false, and in order to prove it is false you need to give a counterexample.

A counterexample would be the function you gave, as it is discountinous at $x=3$.

Proof: $f(3)=5$.

$\lim_\limits{x\to 3}f(x)=\lim_\limits{x\to 3}\frac{x^3-3x^2}{(x-3)}=\lim_\limits{x\to 3}\frac{3x^2-6x}{1}$ (L'Hôpital's rule)$\implies\lim_\limits{x\to 3}f(x)=9\neq5$.

Answer 3

Any discontinuous function disproves the statement, by definition of continuity. A single example suffices.

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If you look at the definition of a limit, you will notice that it does not use the value of the function at the targeted point. Hence, that value is completely independent of the limit.