Which power of x wins, near 0 and far away?

1. 1. Close to 0

   Here are $y = x^2$ and $y = x^4$. The view always stretches its height to fit the bigger of the two at its edge, so a curve lying flat along the axis is tiny beside the other there. The slider zooms in and out in powers of ten.

   Question: Zoom in until the view reaches only $x = \pm 0.1$. Which curve will lie flat?

   $y = x^4$. At $x = 0.1$, $x^2 = 0.01$ but $x^4 = 0.0001$, a hundred times smaller. Squaring a number between $0$ and $1$ makes it smaller, so close to $0$ the higher power is the smaller one.

   The figure: The curves y = x² and y = x⁴ out to x = ±0.100, with the view's height fitted to the bigger one. At the edge x² is 0.0100 and x⁴ is 0.000100.

2. 2. Where they swap

   Out at $x = \pm 10$ it is the other way round: $x^4 = 10\,000$ fills the view and $x^2 = 100$ lies flat. Somewhere between $0.1$ and $10$ the two change places.

   Zoom until the two curves meet at the edges of the view.

   They meet at $x = \pm 1$, where $x^2 = x^4 = 1$. Inside that, $x^4 < x^2$; outside it, $x^4 > x^2$. Any two powers of $x$ swap sizes there: near $0$ the lower power is bigger, far away the higher one.

   $$x^4 < x^2 \text{ for } 0 < |x| < 1, \qquad x^4 > x^2 \text{ for } |x| > 1$$$x^4$ $<x^2\text{ for }0$ $<|x|$ $<1,$$x^4$ $>x^2\text{ for }|x|$ $>1$

   The figure: The curves y = x² and y = x⁴ out to x = ±1.00, with the view's height fitted to the bigger one. At the edge x² is 1.00 and x⁴ is 1.00.

3. 3. A sum near 0

   Now $f(x) = x^3 - 4x$, with its two terms dashed. To sketch $f$ near $0$ you need the term that is bigger there. A common slip picks the higher power and draws $f$ near $0$ like the amber curve, $y = x^3$.

   Question: Zoom in to $x = \pm 0.1$. What will the graph of $f$ look like there?

   Like $y = -4x$. At $x = 0.1$ the $-4x$ term is $-0.4$ and the $x^3$ term only $0.001$, $400$ times smaller, so $f$ runs along the line while the amber curve lies flat on the axis. Near $0$ the lower power sets the shape.

   The figure: The graph of f(x) = x³ − 4x out to x = ±0.100, with its terms y = x³ and y = −4x dashed, y = x³ marked as the slip. At the edge x³ is 0.00100, −4x is −0.400 and f is −0.399.

4. 4. Far from 0

   At $x = 2$ the two terms are the same size, $8$, and cancel: $f(2) = 0$. Beyond $2$ the $x^3$ term is the bigger one, and by more the further out you go.

   Zoom out until $f$ is within $1\%$ of its $x^3$ term at the edge of the view.

   At $x = 20$, $x^3 = 8000$ and $f(20) = 7920$, $99\%$ of it. Far away $f$ looks like $x^3$ and near $0$ like $-4x$: with its zeros at $0$ and $\pm 2$, that is enough to sketch it.

   $$x^3 - 4x \approx -4x \text{ near } 0, \qquad x^3 - 4x \approx x^3 \text{ far from } 0$$$x^3-4x$ $\approx -4x\text{ near }0,$$x^3-4x$ $\approx x^3\text{ far from }0$

   The figure: The graph of f(x) = x³ − 4x out to x = ±20.0, with its terms y = x³ and y = −4x dashed. At the edge x³ is 8000, −4x is −80.0 and f is 7920.

5. 5. Beyond every power

   Here are $y = e^x$ and $y = x^5$ out to $x = 10$. There $x^5 = 100\,000$, more than four times $e^{10} \approx 22\,026$, and $e^x$ is the curve that looks small.

   Question: Zoom out to $x = 20$. Which is bigger at $x = 20$?

   $e^x$: $e^{20} \approx 4.85 \times 10^8$ against $20^5 = 3.2 \times 10^6$, about $150$ times bigger. It overtook $x^5$ near $x = 12.7$. One more unit of $x$ multiplies $e^x$ by $e \approx 2.72$ wherever you are, but $x^5$ by only $(21/20)^5 \approx 1.28$ at $x = 20$. Far enough out, $e^x$ beats every power.

   The figure: The curves y = eˣ and y = x⁵ from x = 0 out to x = 20.0, with the view's height fitted to the bigger one. At the edge eˣ is 4.85 × 10⁸ and x⁵ is 3.20 × 10⁶.

The film, as said, 1:29

1. 0:00

   Here are $y = x^2$ and $y = x^4$. The view stretches its height to fit the bigger one at its edge.

2. 0:06

   Zoom in to $x = \pm 0.1$. Now $x^4$ lies flat: $0.0001$ against $0.01$, a hundred times smaller.

3. 0:14

   Squaring a number between $0$ and $1$ shrinks it, so near $0$ the higher power is the smaller.

4. 0:20

   Zoom out to $x = \pm 10$, and they swap: $x^4 = 10\,000$, and $x^2$ is only $100$.

5. 0:29

   They change places at $x = \pm 1$, where $x^2 = x^4 = 1$.

6. 0:34

   Now $f(x) = x^3 - 4x$, its terms dashed. A common slip sketches it near $0$ like the amber $y = x^3$.

7. 0:41

   Zoom in to $x = \pm 0.1$: $f$ runs along $y = -4x$, and the amber curve lies flat.

8. 0:48

   There $-4x$ is $-0.4$ and $x^3$ only $0.001$. Near $0$ the lower power sets the shape.

9. 0:54

   At $x = 2$ the two terms are both $8$ in size, and cancel: $f(2) = 0$.

10. 1:01

    Out at $x = 20$, $f(20) = 7920$, within $1\%$ of $x^3 = 8000$. Far away, $f$ looks like $x^3$.

11. 1:09

    Last, $y = e^x$ against $y = x^5$. At $x = 10$, $x^5$ is more than four times $e^{10}$.

12. 1:15

    Zoom out to $x = 20$: $e^x$ overtook $x^5$ near $x = 12.7$, and is now about $150$ times bigger.

13. 1:22

    So near $0$ the lower power wins, far away the higher, and far enough out $e^x$ beats every power.