## Question

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

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

## Answers

#lim_(x->oo) sinx/x = 0# ## Explanation:

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!

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