Lim x -> inf. of sinx/x

[PhizKid]

Homework Statement

$$\lim_{x \to \infty } \frac{\sin x}{x}$$

Homework Equations

The Attempt at a Solution

$$\lim_{x \to \infty } \frac{\sin x}{x}$$
Substitute x = 1/u
$$\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1$$

I know this is incorrect because intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence.

 

[Dick]

Homework Statement

$$\lim_{x \to \infty } \frac{\sin x}{x}$$

Homework Equations

The Attempt at a Solution

$$\lim_{x \to \infty } \frac{\sin x}{x}$$
Substitute x = 1/u
$$\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1$$

I know this is incorrect because intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence.

Translate what you correctly know into an argument using the squeeze theorem.

 

[PhizKid]

We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)

 

[Dick]

We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)

No, there's no need to memorize it. Remember -1<=sin(x)<=1, so -1/x<=sin(x)/x<=1/x. Even if you don't have the theorem that should make it pretty obvious that the limit is 0.

 

[Mentallic]

$$\lim_{x \to \infty } \frac{\sin x}{x}$$
Substitute x = 1/u
$$\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1$$

That doesn't look right, if you're going to plug 1/u in for x then everything else needs to stay the same. As you have it there, you've changed the limit from $x\to\infty$ to $\frac{1}{u} \to 0$ when it should be $\frac{1}{u} \to \infty$
But regardless, this won't help you.

We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)

Well then I guess you can just memorize it, but the squeeze theorem isn't very hard to understand really, especially for this example.

It's true that $$-1\leq \sin(x)\leq 1$$ right?
So if we divide all through by x, we get $$\frac{-1}{x}\leq \frac{\sin(x)}{x}\leq \frac{1}{x}$$
And now we know that the limit as $x\to\infty$ for both $-1/x$ and $1/x$ is equal to zero, and since $\sin(x)/x$ is in between those, its limit must also be zero.

 

[PhizKid]

Wow, thank you. That helps a ton, so now I know how to explicitly state this