How do you find the limit of $\frac{sin x}{x}$ $sinx / x$ as x approaches $\infty$ $oo$ ?

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How do you find the limit of $\frac{sin x}{x}$ $sinx / x$ as x approaches $\infty$ $oo$ ?

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Answer 1 · 2016-03-29T13:11:33

We use squeeze theorem, which says that for three functions $g (x), f (x), and h (x)$ $g(x), f(x), and h(x)$ ,
If $g(x)<= f(x) <= h(x), and lim_ (x->a)g(x)=lim_ (x->a)h(x)=L$
then $lim_ (x->a)f(x)=L$ .

$lim_(x->+-oo)sin x/x=0$

Explanation:

Given that $f(x)=sinx/x$
We know that $sinx$ oscillates between $-1 and +1$ for all values of $x$ . Therefore, if we set $g(x)=(-1)/xand h(x)=(+1)/x$ we have located the two functions satisfying the first condition that
$g(x)<= f(x) <= h(x)$ which can be written as

Given that $(-1)/x<= sinx/x <= (+1)/x "for all values of " x " in " (-oo,oo)$
We know that $lim_(x->+-oo)(-1)/x=0 and "also that" lim_(x->+-oo)(+1)/x=0$ .

It follows from the squeeze theorem that
$lim_(x->+-oo)sin x/x=0$

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How do you find the limit of $\frac{sin x}{x}$ $sinx / x$ as x approaches $\infty$ $oo$ ?

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How do you find the limit of $\frac{sin x}{x}$ $sinx / x$ as x approaches $\infty$ $oo$ ?