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I've just started learning about limits. Why can we say $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 0 $$ even though $\lim_{x\rightarrow \infty} \sin x$ does not exist?

It seems like the fact that sin is bounded could cause this, but I'd like to see it algebraically.

$$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = \frac{\lim_{x\rightarrow \infty} \sin x} {\lim_{x\rightarrow \infty} x} = ? $$

L'Hopital's rule gives a fraction whose numerator doesn't converge. What is a simple way to proceed here?

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    $\begingroup$ note that $|\sin(x)|\le 1$ for all real $x$ $\endgroup$ – Dr. Sonnhard Graubner Jul 20 '15 at 21:59
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    $\begingroup$ you could use the $\epsilon$-$\delta$ definition of convergence and see it works... $\endgroup$ – hjhjhj57 Jul 20 '15 at 22:00
  • $\begingroup$ Don't know much about analisis but since $|sin(x)|$ is bounded and $\frac{1}{x} \rightarrow 0$, then $lim_{x \rightarrow +\infty} \frac{sin(x)}{x}=0$ $\endgroup$ – Leafar Jul 20 '15 at 22:01
  • $\begingroup$ Why algebraically? It is clear that when $x$ is very big, then $\frac{\sin x}{x}$ is close to $0$. How close? If $x=100000$, we have $\left|\frac{\sin x}{x}=0\right|\le 10^{-5}$. $\endgroup$ – André Nicolas Jul 20 '15 at 22:01
  • $\begingroup$ @AAron, please do not use MathJax only titles... I've edited the post to fix your mistake this time. $\endgroup$ – Zain Patel Jul 20 '15 at 22:39
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You know that $-\dfrac{1}{x} \leq \dfrac{\sin(x)}{x} \leq \dfrac{1}{x}$

Now let $x \to \infty$ and apply the squeeze theorem.

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  • $\begingroup$ Oh, really? What about $x = \frac{3\pi}{2}$? $\endgroup$ – Zain Patel Jul 20 '15 at 22:00
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    $\begingroup$ The lower bound is not true (though it would be with $-1/x$ instead). $\endgroup$ – Ian Jul 20 '15 at 22:00
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    $\begingroup$ Corrected. Clearly too fast when typing. $\endgroup$ – Lundborg Jul 20 '15 at 22:00
  • $\begingroup$ Have you tried plugging negative numbers into your inequality? $\endgroup$ – Matt Samuel Jul 22 '15 at 2:34
  • $\begingroup$ Thats not really relevant when $x \to \infty$ though is it? But ofc you're right the inequality only holds for $x>0$. $\endgroup$ – Lundborg Jul 22 '15 at 11:19
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If the limit of the numerator and denominator both existed, you could take the quotient of the limits. But they don't.

If the limit of the numerator and denominator were both $0$ or both infinite, then you could use L'Hospital's rule. But in fact the limit of the numerator doesn't exist at all.

Instead, this is easier to explain with the definition:

$$\left | \frac{\sin(x)}{x} - 0 \right | \leq \frac{1}{|x|}$$

so given $\varepsilon > 0$, if $x \geq \frac{1}{\varepsilon}$ then $\left | \frac{\sin(x)}{x} \right | \leq \varepsilon$. In other words you can make the ratio arbitrarily small in magnitude by taking $x$ sufficiently large.

If you prefer, you can invoke the squeeze theorem with the lower bound $-1/x$ and the upper bound $1/x$.

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We have $$\lim\frac{f}{g}=\frac{\lim f}{\lim g} $$ if all three limits exist and $\lim g$ is not zero. This doesn't mean that the existence of th elimit on the left would imply the existence of both limits on the right!

Similarly, note for the application of l'Hopital that some conditions must be met: The limit $\lim\frac fg$ must be an "indeterminate form" and the limit of $\lim\frac{f'}{g'}$ must exist; neither is the case here.

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We know that $\sin(x)$ will always be between $[-1,1]$ for $x \in \mathbb{R}$.

The denominator tends to $+\infty$ as $x \to \infty$. So we can conclude, since any member of $\mathbb{R}$ in the interval $[-1,1]$ divided by $+\infty$ that the limit is: $$\frac{x\in[-1,1]}{\infty} = \boxed0$$

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From the boundedness of sinusoidal curves we know that

$$|\sin x \ | \le 1$$ hence $$\bigg|\frac{\sin x}{x}\bigg| \leq \frac{1}{x}$$ Using our rules for absolute inequalities, we find that $$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$$

Now $$\lim_{x \to \infty} -\frac{1}{x} = 0 = \lim_{x \to \infty} \frac{1}{x} $$

If you are unsure as to why this is true, consider the following plot of $\displaystyle f(x) = \frac{1}{x}$. The same argument for $\displaystyle f(x)= - \frac{1}{x}$, since this is simply a reflection in the $x$-axis.

(Note: I plotted it only for positive $x$)

enter image description here

Thus, from the Squeeze Theorem, we thus have that $$\lim_{x \to \infty}\frac{\sin x}{x}=0$$

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$\sin(x)$ is never undefined

and $\frac{\text{any number}}{\infty}=0$

Therefore you can deduce that it comes out to $0$

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  • $\begingroup$ What about $\frac{x}{x}$? "$x$ is never undefined, and $\frac{\text{any number}}\infty=0$." (I see what you're trying to say, but I think you might want to emphasize that this only works because the numerator of $\frac{\sin x}x$ doesn't go to infinity.) $\endgroup$ – Akiva Weinberger Jul 21 '15 at 3:02

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