Suppose that $L$ is a real number and $f$ is a real-valued function defined on some interval $(b, \infty)$. We say that $\displaystyle{\lim_{x \to \infty} f(x) =L}$ if for every positive real number $\epsilon$, there is a real number $M$ such that if $x>M$ then $|f(x) -L| < \epsilon$.
Is this statement correct, or should it be amended to imply that a limit can exist at L (i.e. it is possible for a limit to exist at L), but does not have to be the limit of the function? For example, we can prove from this definition that $\displaystyle{\lim_{x \to \infty} \frac{4}{x^2}=0}$, but can't one also prove that $\displaystyle \lim_{x \to \infty} \frac{4}{x^2}=-0.001$, $\displaystyle \lim_{x \to \infty} \frac{4}{x^2}=-0.0001$ and other false claims by application of this definition?
Answer
The statement is correct.
(Note also that we usually talk about a limit existing at $a$ to refer to the point that the variable $x$ is approaching, rather than what the values of the function are approaching; to refer to what the function is approaching, we talk about the limit being $L$, or equaling $L$).
In your example, you cannot prove that $\lim\limits_{x\to\infty}\frac{4}{x^2} = -.0001$: given any $L\gt 0$, let $\epsilon = \frac{L}{2}$. Then for any $N\gt 0$, pick $x\gt\max\{N, \sqrt{\frac{8}{L}}\}$. Then
$$\frac{8}{L}\lt x^2,\text{ therefore }\frac{4}{x^2}\lt\frac{L}{2}.$$
And therefore, we have that
$$\left|L-\frac{4}{x^2}\right| = L - \frac{4}{x^2} \gt L-\frac{L}{2} = \epsilon.$$
We have therefore proven that if $L\gt 0$, then:
For every $N\gt 0$ there exists $x\gt N$ such that $|L-f(x)|\gt\frac{L}{2}$.
This proves that the limit definition cannot be satisfied, since the condition fails for at least one $\epsilon$.
If $L\lt 0$, pick $\epsilon=\frac{-L}{2}$ and a similar computation shows that you can always find $x$ greater than any given $N$ that will show the property is not satisfied.
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