# What is the value for $\lim_{x\to\infty} \frac{\sin x}{x}$? [duplicate]

What is the value for $\lim \limits _{x\to\infty} \frac{\sin x} x$?

I solved it by expanding $\sin x$ as

$$\sin x = x - \frac {x^3} {3!} \dotsc$$

So $\lim \limits _{x\to\infty} \frac {\sin x} x = 1 -\infty = - \infty$,

but the answer is $0$. Why? What I am doing wrong?

## marked as duplicate by Daniel FischerJan 20 '16 at 19:29

• Use squeeze theorem via $\frac{-1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$. The specific error is that the Taylor series has both positive and negative terms, so you should be getting $1-\infty+\infty$, which is indeterminate. – vadim123 Jan 10 '16 at 16:23
• The immediate reason you're mistaken is that $\sin(x)$ also equals $x - \frac{x^3}3 + \frac{x^5}{720} - \cdots$, so it's $1 - \infty + \infty - \cdots$, which cannot be decided using that approach. – Arthur Jan 10 '16 at 16:23
• Related question: math.stackexchange.com/questions/857532/… – Martin Sleziak Jan 20 '16 at 17:59

Yes , the answer is $0$ .
$$\left |\frac{\sin x}{x}\right | \leq \frac{1}{x}$$ when $x>0$ (this happens because $|\sin x\ | \leq 1$ )
When $x \to \infty$ we have $\frac{1}{x} \to 0$ so the limit must be $0$ .
The range of $\sin(x)$ will always be a value between -1 and 1, no matter what the input. However, there is no such restriction on the denominator. Therefore, if your numerator is restricted to a finite value, and your denominator is not, as the denominator goes to infinity the value of the whole expression will go to zero.