Find the limit as x approaches infinity of $y=sinx/x$ ?

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Answer 1 · 2018-03-21T15:06:03

$lim_(x->oo) sinx/x = 0$

You're going to want to use the squeeze theorem for this. Recall that $sinx$ is only defined on $-1 ≤ sinx ≤ 1$ . Therefore

$-1/x ≤ sinx/x ≤ 1/x$

And since $lim_(x-> oo) -1/x = lim_(x-> oo) 1/x = 0$ , then $lim_(x->oo) sinx/x = 0$ .

Hopefully this helps!

Find the limit as x approaches infinity of y=sinx/xy=sinx/x?