2
$\begingroup$

So for $\lim_{x \to \infty}f(x)=\infty$ by definition means:

$\forall M>0 \exists N$ such that $\forall x\geq N, f(x)\geq M$

The definition for $\lim_{x \to -\infty}f(x)=-\infty$ kind of confuses me. I saw from many sources they use different notation and I was wondering if both of them are correct; here they are:

$\forall (-M)>0 \exists N$ such that $\forall x\leq N, f(x)\leq -M$ and I was wondering if this is equivalent to:

$\forall M<0 \exists N$ such that $\forall x\leq N, f(x)\leq M$. Please tell me thanks!!

The second one makes more sense to me but I'm not sure if its correct.

$\endgroup$
1
$\begingroup$

To define $\lim_{x \to -\infty}f(x)=-\infty$ you want a statement that says something like "as $x$ approaches negative infinity, $f(x)$ is always smaller than any given negative number". Equivalently, given any arbitrarily large negative number $M < 0$, there exists an $N < 0$ such that $f(x) \leq M$ for all $x \leq N$. This is the same as the second definition you have stated.

The first definition you state is slightly incorrect though, it should read

$$\forall M>0 \; \exists N>0 \; s.t. \; \forall x\leq -N, f(x)\leq -M$$

This says that given any arbitrarily large negative number, we can always make $x$ large enough in the negative direction so that $f(x) \leq -M$. This is certainly equivalent to the second definition you have stated.

To see that the first definition doesn't work, notice that if $-M >0$ and $f(x) \leq -M$, then $f$ can take on any non-positive value; it doesn't necessarily have to approach negative infinity as $x$ does.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.