Question

How do you find the limit of `sinx / x` as x approaches `oo`?

Answer 1

We use squeeze theorem, which says that for three functions `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`