# Thread: sinx/x as x goes to infinity

1. ## sinx/x as x goes to infinity

I know Limit goes to 1 as x tends to zero but what happens when x tends to infinity. $\lim_{x \rightarrow infinity} \frac{\sin x}{x}$

2. Originally Posted by charikaar
I know Limit goes to 1 as x tends to zero but what happens when x tends to infinity. $\lim_{x \rightarrow infinity} \frac{\sin x}{x}$

lemma: if $f(x)$ is bounded in some neighborhood of $x_o$ and $g(x) \xrightarrow [x\to x_0] {} 0$, then
$f(x)g(x) \xrightarrow [x\to x_0] {} 0$
If the limit is when $x \rightarrow \infty$ then $f(x)$ must be bounded for $x>R\in \mathbb{R}$, for some real number R

Tonio

3. Hello, charikaar!

I know: . $\lim_{x\to0}\frac{\sin x}{x} \:=\:1$

but what happens when $x$ tends to infinity? . $\lim_{x \to \infty} \frac{\sin x}{x}$

The numerator oscillates between -1 and +1,
. . while the denominator tends to infinity.

Therefore: . $\lim_{x\to\infty}\frac{\sin x}{x} \;=\;0$

4. Well $0\leq |\sin x| \leq 1$, so for $x > 0$,

$0 \leq \frac{|\sin x|}{x} \leq \frac{1}{x}$

By the squeeze theorem, $\frac{\sin x}{x}\rightarrow 0$ as $x \rightarrow \infty$.

Edit : damn, beat me to it.

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