Answer
See explanations.
Work Step by Step
Step 1. Identify the statement in the Exercise: "The number $L$ is the limit of $f(x)$ as $x$ approaches $c$ if, given any $\epsilon\gt0$, there exists a value of $x$ for which $|f(x)-L|\lt\epsilon$
Step 2. Recall the definition of a limit: "We say that $\lim_{x\to c}f(x)=L$ the limit of $f(x)$ as $x$ approaches $c$ is the number $L$, if, for every number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$, $0\lt |x-c| \lt \delta$, we get $|f(x)-L|\lt\epsilon$."
Step 3. Comparing the above two statements, we find that the first statement did not specify all the $x$ values near $c$ but used a single $x$ value instead.
Step 4. An example (which can show the first statement is wrong) is the function $f(x)=sin\frac{1}{x}$ when $x$ approaches $0$. As the function is continuous and oscillates in the range of $[-1,1]$ when $x\to0$, for any small $\epsilon\gt0$, we can find a $x$ value that $|sin\frac{1}{x}-0|\lt\epsilon$. However, the limit $\lim_{x\to0}sin\frac{1}{x}\ne0$ (as a matter of fact, this limit does not exist).
