Why is a 0/0 limit a ratio of slopes?

1. 1. Zoom in on 0/0

   Here are $f(x) = x^3 - 1$ and $g(x) = \log x$, dashed. Both are $0$ at $x = 1$, so $f(1)/g(1)$ is $0/0$: no value at all. A limit as $x \to 1$ is about the points near $1$, so look closely there.

   Question: Zoom in on $(1, 0)$, a thousand times. What will the two curves look like?

   Two straight lines through $(1, 0)$: $f$ with slope $f'(1) = 3$ and $g$ with slope $g'(1) = 1$. Up close a smooth curve is its tangent line, and these two lines have different slopes, so they stay apart however far you zoom.

   The figure: The curves y = x³ − 1 and, dashed, y = log x near (1, 0), magnified 1000 times. Both pass through (1, 0).

2. 2. Heights become slopes

   The probe at $x$, right of $1$, measures both heights: the green bar is $f(x)$, the grey one $g(x)$. Their quotient is under the figure. As you zoom in, the probe closes in on $1$.

   Zoom in until the quotient is within $0.01$ of the ratio of the slopes, $f'(1)/g'(1)$.

   Up close $f(x) \approx 3(x - 1)$ and $g(x) \approx 1\cdot(x - 1)$: each height is a slope times the same small run, and the run cancels. So the $0/0$ quotient tends to the ratio of the slopes, $3/1 = 3$. That is l'Hôpital's rule.

   $$f(a) = g(a) = 0 \quad\Longrightarrow\quad \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{when the limit on the right exists}$$$f(a)$ $=g(a)$ $=0$$⟹$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$ $=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$\text{when the limit on the right exists}$

   The figure: The curves y = x³ − 1 and, dashed, y = log x near (1, 0), magnified 316 times. Both pass through (1, 0). At the probe, x = 1.00095, the top is 0.00285 and the bottom 9.48 × 10⁻⁴: quotient 3.004.

3. 3. Not 0/0

   Lift the top by $1$: now $f(x) = x^3$, so at $x = 1$ the quotient is $1/0$, not $0/0$. Differentiating top and bottom anyway gives $3x^2 / (1/x) = 3x^3$, which tends to $3$, the same as before. The amber curve is the old top, $x^3 - 1$.

   Question: As $x \to 1$ from the right, what does $f(x)/g(x)$ do now?

   It grows without bound: at $x = 1.01$ it is already over $100$. The top stays near $1$ while the bottom shrinks to $0$. Derivatives cannot see a lift: $x^3$ and the amber $x^3 - 1$ have the same slope everywhere, so $3$ is the limit for the amber top, the one through $(1, 0)$.

   The figure: The curves y = x³ − 1 + 1.00 and, dashed, y = log x near (1, 0), magnified 1.4 times. In amber, the unlifted top y = x³ − 1, which passes through (1, 0). At the probe, x = 1.010, the top is 1.03 and the bottom 0.00995: quotient 103.544.

4. 4. Check the sign

   A top heading to $1$ over a bottom heading to $0$ decides the limit with no rule at all, once the bottom's sign is known: $\log x$ is positive right of $1$ and negative left of it.

   Drag the probe to the left of $1$, within $0.05$ of it.

   At $x = 0.99$ the quotient is about $-97$, heading to $-\infty$; from the right it heads to $+\infty$. So there is no limit, and certainly not $3$. Check the form first: the rule is for $0/0$ and $\infty/\infty$ only, and a quotient heading anywhere else decides itself.

   The figure: The curves y = x³ − 1 + 1.00 and, dashed, y = log x near (1, 0), magnified 1.4 times. In amber, the unlifted top y = x³ − 1, which passes through (1, 0). At the probe, x = 0.990, the top is 0.970 and the bottom −0.0101: quotient −96.544.

The film, as said, 1:42

1. 0:00

   Here are $f(x) = x^3 - 1$ and $g(x) = \log x$, dashed. Both are $0$ at $x = 1$, so there the quotient is $0/0$.

2. 0:06

   So look closer. Magnify the point ten times.

3. 0:11

   A hundred times, then a thousand. Both curves straighten into lines through $(1, 0)$.

4. 0:20

   Their slopes are $f'(1) = 3$ and $g'(1) = 1$. Up close, each curve is its tangent line.

5. 0:25

   Step back and add a probe at $x$, right of $1$. Its green bar is $f(x)$, its grey bar $g(x)$.

6. 0:34

   Zoom in, and the probe closes in on 1 with the view. Watch the quotient under the figure.

7. 0:44

   Up close the heights are about $3(x - 1)$ and $1 \cdot (x - 1)$. The run cancels, and the quotient tends to $3$.

8. 0:50

   That is l'Hôpital's rule: when top and bottom are both $0$ at $a$, the limit of $f/g$ is the limit of $f'/g'$.

9. 0:57

   Zoom back out, for a top that does not reach 0.

10. 1:04

    Lift the top by $1$, to $x^3$. At $x = 1$ the quotient is now $1/0$, not $0/0$.

11. 1:11

    The rule cannot see a lift: $x^3$ has the same slopes as the amber $x^3 - 1$, so it still answers $3$.

12. 1:18

    But the quotient grows without bound: at $x = 1.01$ it is already over $100$.

13. 1:25

    Left of $1$, $\log x$ is negative: at $x = 0.99$ the quotient is about $-97$, heading to $-\infty$.

14. 1:33

    Check the form first: the rule is for $0/0$ and $\infty/\infty$. Where top and bottom both reach $0$, their quotient becomes the ratio of their slopes.