I am studying limits at infinity, and I have a doubt about evaluating them. From what I know, limits only exist if both sides of the limit exist and are equal. For example, take a look at the following limit:
$$\lim_{ x\to \infty}\frac{x}{e^{-x}}$$
As $x \to +\infty$ the function goes to $+\infty$. However, as $x \to -\infty$ the function goes to $0$.
Does this mean that the limit does not exist or I only have to evaluate the limit as x goes to $+\infty$?
Best Answer
Although closely related:
i) $\lim_{x\rightarrow a}f(x) = L; a\in \mathbb R$
ii) $\lim_{x\to a^+}f(x) = L$
iii) $\lim_{x\to a^-}f(x) = L$
iv) $\lim_{x\to \infty} f(x) = L$
and v) $\lim_{x\to -\infty} f(x) = L$.
have technically different definitions.
i) means as $x$ gets close to $a$, then $f(x)$ gets close to $L$. or technically:
i) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $|x -a| < \delta$ it will follow that $|f(x) - L| < \epsilon$.
ii) means as $x$ gets close to $a$ but larger than $a$, then $f(x)$ gets close to $L$. or technically:
ii) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $a < x < a+ \delta$ it will follow that $|f(x) - L| < \epsilon$.
Notice that if i) is true than ii) must be true but if ii) is true i) is not nescessarily true.
iii) means as $x$ gets close to $a$ but smaller than $a$, then $f(x)$ gets close to $L$. or technically:
iii) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $a -\delta < x < a$ it will follow that $|f(x) - L| < \epsilon$.
Notice that if i) is true than ii) and iii) must be true. And if both ii) and iii) is true then i) is true. But if one or the other of ii) or iii) is true but the other isn't i) will not be true.
So the statement "i) if ii) and iii)" is not a definition but an observation. Well.... it could be a definition as they are equivalent.
Now...
Limit to infinity is not that as $x$ gets close to infinity $f(x)$ will get close to $L$. That wouldn't make any sense because we can't get $x$ "close to infinity". Instead we mean, as $x$ becomes large, $f(x)$ gets close to $L$. or technically:
iv) For any $\epsilon > 0$ there is some $M \in \mathbb R$ so that whenever $x > M$ it will follow that $|f(x) - L| < \epsilon$.
It makes no sense to talk of $\lim_{x\to \infty^+} f(x)$ because we can't "approach from the right". And to talk of $\lim_{x \to \infty^-}f(x)$ is the exact same thing as $\lim_{x \to \infty}f(x)$ because we can only "approach from the left".
So it is NOT the case that $\lim_{x\to\infty} f(x) = L$ if $\lim_{x\to \infty^+}f(x) = L$ and $\lim_{x\to \infty^-}f(x) = L$. Instead it is just $\lim_{x\to\infty} f(x) = L$ if $\lim_{x\to \infty^-}f(x) = L$.
$\lim_{x \to -\infty}f(x)=L$ means we are talking the values of very big negative numbers (very small, or negative with very large magnitude or absolute values). It should be very clear that $-\infty$ means infinity beyond the negative extreme of the reals, whereas $\infty^-$ means something entirely different; it means approaching the the positive extreme of the real numbers but at some finite value less than infinity.
v) For any $\epsilon > 0$ there is some $M \in \mathbb R$ so that whenever $x < M$ it will follow that $|f(x) - L| < \epsilon$.
Again it makes no sense to talk of $\lim_{x\to -\infty^-}f(x)$ and it is redundant to talk of $\lim_{x\to -\infty^+} f(x)$ for the exact same reasons.