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This question already has an answer here:

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?

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marked as duplicate by Daniel Fischer Jan 20 '16 at 19:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ 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. $\endgroup$ – vadim123 Jan 10 '16 at 16:23
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    $\begingroup$ 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. $\endgroup$ – Arthur Jan 10 '16 at 16:23
  • $\begingroup$ Related question: math.stackexchange.com/questions/857532/… $\endgroup$ – Martin Sleziak Jan 20 '16 at 17:59
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Yes , the answer is $0$ .

One way to see this is by using the inequality :

$$\left |\frac{\sin x}{x}\right | \leq \frac{1}{x}$$ 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 $0$ .

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  • $\begingroup$ I got the point . Thank you ! $\endgroup$ – Yogus Jan 10 '16 at 16:32
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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.

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