Proving $\infty$ Limit of $x\sin(\frac{\pi}{x})=1$

[Karol]

Homework Statement

Prove $~\displaystyle \lim_{x \to \infty}\left(x\sin\left(\frac{\pi}{x}\right)\right)=1$

Homework Equations

$$\lim_{x \to 0} \frac{\sin\pi}{x}=0$$

The Attempt at a Solution

If i could multiply $~x\sin\left(\frac{\pi}{x}\right)~$ with something that would cancel the sin
or to express: $~\sin\left(\frac{\pi}{x}\right)=2\sin\left(\frac{\pi}{2x}\right)\cos\left(\frac{\pi}{2x}\right)$
But it doesn't help

 

[fresh_42]

What expressions of the sine function are you allowed / willing to use, e.g. Taylor series or with the help of the exponential function? Have you considered to substitute $x \mapsto \frac{1}{y}$?

 

[Ray Vickson]

Homework Statement

Prove $~\displaystyle \lim_{x \to \infty}\left(x\sin\left(\frac{\pi}{x}\right)\right)=1$

Homework Equations

$$\lim_{x \to 0} \frac{\sin\pi}{x}=0$$

The Attempt at a Solution

If i could multiply $~x\sin\left(\frac{\pi}{x}\right)~$ with something that would cancel the sin
or to express: $~\sin\left(\frac{\pi}{x}\right)=2\sin\left(\frac{\pi}{2x}\right)\cos\left(\frac{\pi}{2x}\right)$
But it doesn't help

The stated result is false.

 

[SammyS]

Homework Statement

Prove $~\displaystyle \lim_{x \to \infty}\left(x\sin\left(\frac{\pi}{x}\right)\right)=1$

Homework Equations

$$\lim_{x \to 0} \frac{\sin\pi}{x}=0$$

The Attempt at a Solution

If i could multiply $~x\sin\left(\frac{\pi}{x}\right)~$ with something that would cancel the sin
or to express: $~\sin\left(\frac{\pi}{x}\right)=2\sin\left(\frac{\pi}{2x}\right)\cos\left(\frac{\pi}{2x}\right)$
But it doesn't help

The relevant equation you give isn't much help. $\ \sin(\pi)=0 \ , \$ so that's saying that lim (0) = 0 .

As Ray says, $\displaystyle \ \lim_{x \to \infty}\left(x\sin\left(\frac{\pi}{x}\right)\right)\ne 1\ .$

However this limit does exist.

You might consider:$\ \displaystyle \lim_{x \to \infty}\left(\frac{\sin\left(\frac{\pi}{x}\right)} {\frac 1 x } \right)$

Added in Edit:
Better yet:
For $\ \displaystyle \left| \frac{\pi}{x}\right | \ll 1 \ ,\$ what is the behavior of $\ \displaystyle \sin\left( \frac{\pi}{x}\right ) \ ?$

 

[Karol]

I made a mistake, i have to prove $~\displaystyle \lim_{x \to \infty}\left(x\sin\left(\frac{\pi}{x}\right)\right)=\pi$
It's about finding the circumference of a circle by approximation of n sided polygon.
$$\lim_{x \to \infty} \left( x \sin \left( \frac{\pi}{x} \right) \right)=\lim_{x \to \infty}\left(x \cdot \left( \frac{\pi}{x} \right) \right)=\pi$$

 

[Mark44]

I made a mistake, i have to prove $~\displaystyle \lim_{x \to \infty}\left(x\sin\left(\frac{\pi}{x}\right)\right)=\pi$
It's about finding the circumference of a circle by approximation of n sided polygon.
$$\lim_{x \to \infty} \left( x \sin \left( \frac{\pi}{x} \right) \right)=\lim_{x \to \infty}\left(x \cdot \left( \frac{\pi}{x} \right) \right)=\pi$$

L'Hopital's Rule works for this problem. You need to rewrite the limit expression in this form:
$$\lim_{x \to \infty}\frac{\sin(\pi/x)}{1/x}$$

 

[Karol]

But the book didn't teach Lhopital's yet, but i derived and it gives π.
Is my previous answer correct, i mean can i make the transition $~\displaystyle \lim_{x \to \infty} \left( x \sin \left( \frac{\pi}{x} \right) \right)=\lim_{x \to \infty}\left(x \cdot \left( \frac{\pi}{x} \right) \right) ~$?

 

[Mark44]

But the book didn't teach Lhopital's yet, but i derived and it gives π.

Without using L'Hopital's Rule, you can do this:
$\frac{\sin(\pi/x)}{1/x} = \frac{\sin(\pi/x)}{\pi/x} \cdot \pi$
By letting $u = \pi/x$, you can take the limit as u approaches zero, and use the fact that $\lim_{u \to 0}\frac{\sin(u)} u = 1$

Is my previous answer correct, i mean can i make the transition $~\displaystyle \lim_{x \to \infty} \left( x \sin \left( \frac{\pi}{x} \right) \right)=\lim_{x \to \infty}\left(x \cdot \left( \frac{\pi}{x} \right) \right) ~$?

That's probably OK, maybe. You can make the argument that if w is close to 0, $\sin(w) \approx w$. The way I outlined above is better, I think.

 

[Ray Vickson]

But the book didn't teach Lhopital's yet, but i derived and it gives π.
Is my previous answer correct, i mean can i make the transition $~\displaystyle \lim_{x \to \infty} \left( x \sin \left( \frac{\pi}{x} \right) \right)=\lim_{x \to \infty}\left(x \cdot \left( \frac{\pi}{x} \right) \right) ~$?

You don't need l'Hospital to get $\lim_{t \to 0} \sin(t)/t = 1$. In fact, you need a demonstration of this result in order to develop expressions for the derivatives of the trigonometric functions.

 

[Karol]

You don't need l'Hospital to get $\lim_{t \to 0} \sin(t)/t = 1$

That is right, they didn't use l'Hopital.
Thank you Ray, Mark, Sammy and fresh.

 

[Ray Vickson]

That is right, they didn't use l'Hopital.
Thank you Ray, Mark, Sammy and fresh.

l'Hopital or l'Hospital? (Pronounced as lo-pee-tal in any case).

The remarks following remarks from Paul's On-line Math Notes deal with this issue:

"Before proceeding with examples let me address the spelling of “L’Hospital”. The more modern spelling is “L’Hôpital”. However, when I first learned Calculus my teacher used the spelling that I use in these notes and the first textbook that I taught Calculus out of also used the spelling that I use here.

Also, as noted on the Wikipedia page for L’Hospital's Rule,

“In the 17th and 18th centuries, the name was commonly spelled "l'Hospital", and he himeself spelled his name that way. However, French spellings have been altered: the silent 's' has been removed and replaced with the circumflex over the preceding vowel. The former spelling is still used in English where there is no circumflex.”

So, the spelling that I’ve used here is an acceptable spelling of his name, albeit not the modern spelling, and because I’m used to spelling it as “L’Hospital” that is the spelling that I’m going to use in these notes."