One sided limits with $f(x)$ instead of $x$

Question

Can you have a limit such that as $x$ approaches $+\infty,-\infty,c^{+},c^{-},c$ then $f(x)$ approaches some value $L$ from the top or bottom.

What would the epsilon-delta definition be?

Answer 1

A second answer for a reread and realizing you probably/maybe meant something other than what I thought you meant.

One rereading your question, it seems you are asking that if we can indicate $x \to c$ but $x < c$ (or $x > c$) we can use the notation $x\to c^{-}$ (or $x \to c^+$), then is there similar notation for indicating $f(x)\to L$ but $f(x)< L$ (or $f(x) > L$)? And if so how do you express it as a delta epsilon?

This is somewhat less conventional but I think (someone correct me if I'm wrong) we can writh $\lim_{blah} f(x) = L^+$ to mean it approaches from above and $\lim_{blah} f(x) = f(x)=L^-$ to mean from below.

(But of course approaching $\infty$ must be from below and approaching $-\infty$ must be from above so $\lim f(x) = \infty$ and $\lim f(x) = \infty^-$ mean the same thing and $\lim f(x) =\infty^+$ is nonsense. Similar for $\lim f(x) = -\infty$ meaning $\lim f(x) =-\infty^+$ and $\lim f(x)=\infty^-$ being meaningless.)

Now the delta-epsilon,N-M definition:

A $lim_{x \to blah} f(x) = BLAH$ has two parts.

The "$x$ gets close to $c$" or "$x$ get 'close' to infinity" (which is, of course, meaningless-- $x$ can't get "close" to infity-- instead $x$ gets arbitrarily large) part.

And the "$f(x)$ gets close to $L$" or "$f(x)$ gets 'close' to infinity" (ditto) part.

And the delta-epsilon, N,M way of writing this is

For every RANGE FOR THE OUTPUT VALUE there is a RANGE FOR THE INPUT VALUE where CONDITION FOR THE $x$ IS CLOSE TO $c$ OR INFINITY PART will imply THE $f(x)$ IS CLOSE TO $L$ OR INFINITY PART

where we fill in those capital letters with what we want.

If we want $f(x) \to L$ then RANGE FOR THE OUTPUT VALUE will be "for every $\epsilon > 0$"

If we want $f(x) \to \pm \infty$ then RANGE FOR THE OUTPUT VALUE will be "for every M \in \mathbb R$.

If we want $f(x)\to L$ but we don't care from which direction the THE $f(x)$ IS CLOSE TO $L$ OR INFINITY PART will be "$|f(x) - L| < \epsilon$".

If we want $f(x) \to L^+$ from above which means $f(x) > L$ and $f(x)-L > 0$ then THE $f(x)$ IS CLOSE TO $L$ OR INFINITY PART will be "$0 < f(x) - L < \epsilon$".

If we want $f(x) \to L^-$ from below which means $f(x) < L$ and $f(x)-L < 0$ then THE $f(x)$ IS CLOSE TO $L$ OR INFINITY PART will be "$0 < L-f(x) < \epsilon$".

If we want $f(x) \to \infty$ which means $f(x)$ gets really really large and larger than the $M$ we specified, then THE $f(x)$ IS CLOSE TO $L$ OR INFINITY PART will be "$f(x) > M$.

And If we want $f(x) \to -\infty$ which means $f(x)$ gets really really large in the negative direction and less than the $M$ we specified, then THE $f(x)$ IS CLOSE TO $L$ OR INFINITY PART will be "$f(x) < M$"

......

Similar for the RANGE FOR THE INPUT VALUE and the $x$ IS CLOSE TO $c$ OR INFINITY PART

If $x\to c$ then the RANGE FOR THE INPUT VALUE will be "there exists a $\delta>0$.

And if $x\to \pm \infty$ then RANGE FOR THE INPUT VALUE will be "there exists and $N \in \mathbb R$.

And if we want $x \to c$ but we don't care what direction then the $x$ IS CLOSE TO $c$ OR INFINITY PART will be "whenever $|x-c|< \delta$".

And if we want $x\to c^-$ so $x< c$ and $c-x > 0$ then the $x$ IS CLOSE TO $c$ OR INFINITY PART will be "whenever $0<c-x< \delta$".

And if we want $x\to c^+$ so $x> c$ and $x-c > 0$ then the $x$ IS CLOSE TO $c$ OR INFINITY PART will be "whenever $0<x-c< \delta$".

And if we want $x\to \infty$ then the $x$ IS CLOSE TO $c$ OR INFINITY PART will be "whenever $x > N$"

And if we want $x\to -\infty$ then the $x$ IS CLOSE TO $c$ OR INFINITY PART will be "whenever $x < N$"

=======

So there are $25$ ways to describe a limit

$$\lim\limits_{x\to c:c^+:c^-:\pm \infty} f(x) = L:L^+:L^-:\pm \infty$$ and there are

twenty five ways to write the defintion

For every ($\epsilon > 0; M\in \mathbb R$) there is a ($\delta > 0; N \in \mathbb R$) so that whenever ($|x-c|< \delta; 0< x-c < \delta; 0< c-x < \delta; x > N; x < N$) that would imply ($|f(x)-L| < \epsilon; 0< f(x)-L< \epsilon; 0< L-f(x) < \epsilon; f(x) > M; f(x) < M)$.

Answer 2

How about $y=\frac{2x^2+4x}{x^2+x}$. It has a horizontal asymptote $y=2$ which addresses to $x$ going to $+/-$ infinity part. And the graph has a hole in the graph at $(0,2)$ so that adresses the $c+$ and $c-$ part. So in all cases $L=2$. Does this answer your question? Otherwise I can take my answer off.

Answer 3

"What would the epsilon-delta definition be?"

You wouldn't have just one.

To have $\lim_{x\to \infty} f(x) = L$ we'd need for every $\epsilon$ there will be an $M_1$ so that $x > M_1$ we'd have $|f(x) -L| < \epsilon$.

To have $\lim_{x\to -\infty}f(x) = L$ we'd need for the same $\epsilon$ there will be an $M_2$ so that $x < M_2$ we'd have $|f(x) -L| < \epsilon$.

We can combine both of them be letting $K \ge \max (|M_1|, |M_2|)$ and saying: for every $\epsilon > 0$ there is a $K > 0$ so that $x > K$ or $x < -K$ will imply $|f(x) -L| < \epsilon$.

Or to combine to a single condition. for every $\epsilon > 0$ there is a $K > 0$ so that if $x > K$ (and so $-x < -K$) then $|f(\pm x) -L| < \epsilon$.

To indicate the comin in from opposite sides we need that if $x > K$ then $f(x)- L$ is positive/neg while if $x < -K$ then $f(x)-L$ is neg/positive. (Note we can make $K$ large enough that we only get the "tail ends" of we can assume for all $x < -K$ or $x > K$ are all above of below $L$)

So we can do this be stating that: For all $\epsilon > 0$ there is a $K > 0$ so that if $x > K$ then $0 < |f(\pm x) - L| < \epsilon$ and $\frac {f(x)-L}{f(-x)-L} < 0$.

Okay the limit's of $c^+, c^-$ are similar. For every $\epsilon >0$ there is $\delta > 0$ so there if $ c < x < c+\delta$ and $c-\delta < w < c$ then $|f(x,w)-L| < \epsilon$ and $\frac {f(x)-L}{f(w) - L} < 0$.

Or to combine $x = c + k$ and $w =c-k$ for some $k$ we can say:

Fore every $\epsilon > 0$ there is $\delta > 0$ so that for all $0 < k < \delta$ then $|f(c\pm k) -L|< \epsilon$ and $\frac {f(c+k) -L}{f(c-k)-L} < 0$....(my.. that looks obscene.... that was uninintentional... I swear!)

So...... putting it altogether:

For all $\epsilon > 0$ there is a $K > 0$ and $\delta > 0$ so that for every $x > K$ and for every $k; 0< k< \delta$ we'd have $|f(\pm x) - L| < \epsilon; |f(c\pm k) -L|< \epsilon$ and $\frac {f(x)-L}{f(-x)-L}<0$ and $\frac {f(c+k)-L}{f(c-k)-L} < 0$.

.....

Now you didn't state that for all $x \in (-\infty, c)$ you wanted $f(x)$ to be one side of the $L$ and for all $x \in (c, \infty)$ you wanted $f(x)$ to be on the other side.

If that was a requirement we can say.... this gets wordy....

For all $w < c$ and $x > c$ then $f(w)\ne L$ and $\frac {f(x)-L}{f(w)-L} < 0$ and for all $\epsilon > 0$ there is a $K > 0$ and $\delta > 0$ so that for every $x > K$ and for every $k; 0< k< \delta$ we'd have $|f(\pm x) - L| < \epsilon; |f(c\pm k) -L|< \epsilon$ and $\frac {f(x)-L}{f(-x)-L}<0$ and $\frac {f(c+k)-L}{f(c-k)-L} < 0$

Phwew.