I have the following question, and not sure if its a true or false statement. Couldn't construct a proof but couldn't construct a counter example either...
If $\lim \limits_{x \to a} |f(x)| = L$ for some $L≥0$ then either $\lim \limits_{x \to a} f(x) = L$ or $\lim \limits_{x \to a} f(x) = -L$ .
Any help would be appreciated!
If $L = 0$, the statement is true. This is built into the definition of limit. Because saying $f(x)$ is close to $0$ is the same thing as saying $|f(x) - 0|$ is small, which is the same thing as saying $|f(x)|$ is small, which is the same thing as saying $|f(x)|$ is close to $0$.
If $L > 0$, the statement is false. You can picture a function that is “undecided” between $L$ and $-L$ as a limit. But then if you apply the absolute value, you get a function that has a limit of $L$. And example would be $$ f(x) = \begin{cases} 1 & \text{if $x$ is rational} \\ -1 & \text{if $x$ is irrational} \end{cases} $$ Then $|f(x)| = 1$, so $\lim_{x\to 0}f(x) = 1$. But $\lim_{x\to 0} f(x)$ cannot be $1$ because there are infinitely many irrational numbers arbitrarily close to $0$, and cannot be $-1$ because there are infinitely many rational numbers arbitrarily close to $0$. So $f$ has no limit at $0$.
EDIT re your your question
I'm not sure what you mean by contradiction. If you mean that this example is not a counterexample to the given statement (that is, evidence of the statement's falsehood), remember that $L$ has to be the same throughout the statement.
When we say “The statement ‘If $\lim_{x\to a} |f(x)| = L$, then $\lim_{x\to a} f(x) = L$ or $\lim_{x\to a} f(x) =-L$’ is false,” what we mean is: “there exists a function $f$ such that $\lim_{x\to a}f(x) = L$, but neither $\lim_{x\to a} f(x) = L$ nor $\lim_{x\to a} f(x) = -L$ are true”. Substitute $L$ for $1$ (which is positive) and my function satisfies this. I'm not changing $L$ to $-1$ anywhere; $-L$ is just another way of saying $-1$.