What does it mean for f(x) to approach L?

1. 1. A band and a window

   Here $f(x) = \frac{x^3 - x}{x - 1}$, which is not defined at $x = 1$: the formula gives $\frac{0}{0}$ there. The shaded band is every height within $0.5$ of $2$. A limit asks whether some window around $1$, with $1$ itself left out, keeps the graph inside the band.

   Narrow the window until the graph inside it stays inside the band.

   Any $\delta$ up to about $0.158$ works: in that window $f$ stays between $1.5$ and $2.5$. The point $x = 1$ is left out of every window, so it does not matter that $f(1)$ is not defined.

   The figure: The graph of f(x) = (x³ − x)/(x − 1), with a hole at (1, 2). A band of heights from 1.50 to 2.50. In the window within 0.150 of x = 1, x = 1 left out, f runs from 1.57 to 2.00 on the left and 2.00 to 2.47 on the right, all inside the band.

2. 2. However thin the band

   One band is easy. The real test is a challenge that never stops: make the band thinner and ask again. This one is $2 \pm 0.1$, five times thinner, and the view has zoomed in.

   Question: Will some window still keep the graph inside this band?

   Yes: any $\delta$ up to about $0.033$ works. Whatever band you name around $2$, however thin, a narrow enough window keeps the graph inside it. That, and nothing about $f(1)$, is what it means for the limit to be $2$.

   $$\lim_{x \to 1} \frac{x^3 - x}{x - 1} = 2$$$\underset{x\to 1}{\lim }\frac{x^3-x}{x-1}$ $=2$

   The figure: The graph of f(x) = (x³ − x)/(x − 1), with a hole at (1, 2). A band of heights from 1.90 to 2.10. In the window within 0.030 of x = 1, x = 1 left out, f runs from 1.91 to 2.00 on the left and 2.00 to 2.09 on the right, all inside the band.

3. 3. Not stopped by 0/0

   Putting $x = 1$ into the formula gives $\frac{0}{0}$. A common slip stops there and calls the limit $0$: the amber band. The numbers under the figure are the top and the bottom of the fraction at the window's right edge.

   Question: As the window closes, the top and the bottom both go to $0$. What does their ratio do?

   It settles on $2$. At $x = 1.01$ the top is $0.0203$ and the bottom $0.0100$: the top stays about twice the bottom, because $x^3 - x = (x - 1)(x^2 + x)$. $\frac{0}{0}$ says only that the algebra is unfinished: cancel $x - 1$, and the limit is $1^2 + 1 = 2$.

   The figure: The graph of f(x) = (x³ − x)/(x − 1), with a hole at (1, 2). A band of heights from 1.90 to 2.10, and the slip's band from −0.10 to 0.10. In the window within 0.010 of x = 1, x = 1 left out, f runs from 1.97 to 2.00 on the left and 2.00 to 2.03 on the right, all inside the band.

4. 4. When no band works

   Now the right-hand piece is lowered by $1$: $f(x) = x^2 + x$ for $x < 1$ and $x^2 + x - 1$ for $x > 1$. The band is $0.1$ either side of its centre, and you can drag it up and down.

   Question: Is there a centre for the band where a narrow enough window keeps the graph inside?

   None. Just left of $1$ the values are near $2$, just right of it near $1$, and a band $0.2$ tall cannot hold both, however narrow the window. Each side has a limit of its own, $2$ from the left and $1$ from the right; they disagree, so $\lim_{x \to 1} f(x)$ does not exist.

   $$\lim_{x \to 1^-} f(x) = 2 \neq 1 = \lim_{x \to 1^+} f(x)$$$\underset{x\to 1^{-}}{\lim }f(x)$ $=2$ $\ne 1$ $=\underset{x\to 1^{+}}{\lim }f(x)$

   The figure: The graph of f(x) = x² + x for x < 1 and x² + x − 1 for x > 1, with no point at x = 1. A band of heights from 1.40 to 1.60. In the window within 0.020 of x = 1, x = 1 left out, f runs from 1.94 to 2.00 on the left and 1.00 to 1.06 on the right, not all inside the band.

5. 5. Out at infinity

   A limit at infinity plays the same game with another kind of window: every $x$ past some $N$, in place of every $x$ near a point. Here $f(x) = \frac{2x^2 + 1}{x^2 + 1}$, and the band is $2 \pm 0.1$.

   Drag the window's edge $N$ until the graph past it stays inside the band.

   From $N = 3$ on: $f(3) = 1.9$, and past it the graph only climbs toward $2$. A thinner band needs a larger $N$, but some $N$ always works, so $\lim_{x \to \infty} f(x) = 2$ and $y = 2$ is a horizontal asymptote.

   $$\lim_{x \to \infty} \frac{2x^2 + 1}{x^2 + 1} = 2$$$\underset{x\to \infty }{\lim }\frac{2x^2+1}{x^2+1}$ $=2$

   The figure: The graph of f(x) = (2x² + 1)/(x² + 1), climbing toward 2, with the band from 1.90 to 2.10. Past x = 3.0 its values run from 1.90 up toward 2, all inside the band.

The film, as said, 1:44

1. 0:00

   Here $f(x) = \frac{x^3 - x}{x - 1}$. At $x = 1$ it is $\frac{0}{0}$, so the graph has a hole there.

2. 0:05

   The band is every height within $0.5$ of $2$; the dashed lines mark a window around $1$.

3. 0:11

   Narrow the window to $\delta = 0.15$, and the graph in it, with $x = 1$ left out, stays inside the band.

4. 0:19

   Make the band five times thinner, $2 \pm 0.1$, and zoom in. That window no longer works.

5. 0:27

   Narrow it to $\delta = 0.03$ and the graph is inside again: any $\delta$ up to about $0.033$ works.

6. 0:35

   Every band, however thin, has a window that works. That is what a limit of $2$ means.

7. 0:41

   A common slip stops at $\frac{0}{0}$ and calls the limit $0$: the amber band.

8. 0:48

   Close the window. The top and the bottom both go to $0$, but their ratio settles on $2$.

9. 0:58

   At $x = 1.01$ the top is $0.0203$, about twice the bottom, $0.0100$. Cancel $x - 1$ and the limit is $2$.

10. 1:04

    Lower the right-hand piece by $1$. Now the left side heads for $2$ and the right for $1$.

11. 1:12

    A band $0.2$ tall cannot hold both sides, wherever it sits, however narrow the window.

12. 1:20

    The one-sided limits disagree, so the two-sided limit does not exist.

13. 1:24

    At infinity, the window is every $x$ past some $N$. Here $f(x) = \frac{2x^2 + 1}{x^2 + 1}$, and the band is $2 \pm 0.1$.

14. 1:29

    Past $N = 3$ the graph stays inside: $f(3) = 1.9$, and from there it only climbs toward $2$.

15. 1:37

    So a limit is a promise: name any band around $L$, however thin, and some window keeps the graph inside it.