If $ \lim n\tan{\frac{180°}{n}}= \pi $, is $ \infty \times0 = \pi $?

Question

I derived the volume of a cone using two approaches and compared the results.

First I integrated a circle of radius $r$ over the height $h$ to get the expression: $$V_1=\frac{1}{3}\pi r^2 h$$

Then I considered a polygonal pyramid of infinite sides.

An n-sided polygon with apothem $r$ has an area of: $$A=nr^2\tan{\frac{180°}{n}}$$

Integrating this over the height $h$ gives the expression for the area of the n-sided polygonal pyramid as: $$V_2=\frac{1}{3}n\tan{\frac{180°}{n}}r^2 h$$

Equating $V_1$ and $V_2$ implies that: $$ \lim_{n \to \infty} \left(n\tan{\frac{180°}{n}}\right) = \pi $$

So is it true to say that: $$\infty\tan{\frac{180°}{\infty}} = \pi$$

But: $$\tan{\frac{180°}{\infty}}=0$$

So: $$\infty (0)=\pi$$

Can anyone shed some light on this surprising result?

Answer 1

Equating $V_1$ and $V_2$ implies that: $$ \lim_{n \to \infty} \left(n\tan{\frac{180}{n}}\right) = \pi $$

That is correct. 180 degrees is $\pi$ radians. If you change variables from $n$ to $\theta$ with $\theta = \frac{1}{n}$, you get $$ \lim_{n \to \infty} \left(n\tan{\frac{180}{n}}\right) = \lim_{n \to \infty} \frac{\tan{\pi x}}{x} $$ and that is, in fact $\pi$.

So is it true to say that: $$\infty\tan{\frac{180}{\infty}} = \pi$$

Not at all. The limit of a product of functions is the product of the limits of the functions provided those functions have limits in the first place. Since $\lim_{n\to\infty} n$ does not exist, you cannot apply the limit laws that way.

In short: $\infty$ is not a number. Treating it as a number leads to madness (or, at least, contradiction).

Answer 2

You cannot simply replace expressions that tend to infinity by that same symbol; then you would lose information on how fast something tends to infinity, for example.

Take $a_n=n$, $b_n=2n$. Then clearly, both tend to infinity. But $b_n/a_n=2\to2$, and $a_n/a_n=1\to1$. Both limits would be "$\infty/\infty$", but are still very different. What is key here is the way both expressions tend to infinity. You lose this information when you just replace the sequence with a $\infty$ symbol.

Answer 3

This is not a "result". $\infty$ is not a number. If your argument were correct you could use it this way:

For all positive integers $n$ $$ n \times \frac{1}{n} = 1. $$ Then taking the limit as $n \to \infty$ $$ \infty \times 0 = 1. $$

Answer 4

$\lim_{n \to \infty} (n\tan{\frac{180°}{n}}) = \pi$

Take $\frac{1}{n}= x$, then ${n \to \infty} \implies {x \to 0}$

Also we have $$\lim_{n \to \infty} {\frac{\tan x}{x}} = 1$$

So $\lim_{n \to \infty} (n\tan{\frac{180°}{n}}) = \pi \implies \lim_{x \to 0} ({\frac{\tan 180° x}{180° x}}180°) = \pi \implies 180°=\pi$

Answer 5

Generally speaking, $\infty\cdot0$ has no precise value (not counting that it is a mathematically "illegal" expression).

For more rigor, you can work with limits and write

$$\infty\cdot0=\lim_{n\to \infty}f(n)g(n)$$ where $\lim_{n\to \infty}f(n)=\infty$ and $\lim_{n\to \infty}g(n)=0$.

Then you have the following examples:

• $f(n)=n$ and $g(n)=\tan\dfrac1n\implies \infty\cdot0=1,$

• $f(n)=n^2$ and $g(n)=\tan\dfrac1n\implies \infty\cdot0=\infty,$

• $f(n)=n$ and $g(n)=\tan\dfrac1{n^2}\implies \infty\cdot0=0,$

• $f(n)=n^3$ and $g(n)=\dfrac1n-\tan\dfrac1n\implies \infty\cdot0=-\dfrac13$

• $\cdots$

---

In your case,

$$\tan°\frac{180°}n=\tan\frac\pi n$$ where the second tangent function has its argument expressed in radians. Then for small arguments,

$$\tan\frac\pi n\approx \frac\pi n$$

so that

$$n\tan\frac\pi n\approx \pi$$ and this is exact in the limit.