# `lim_(x->oo)cosx/x` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve ,

`lim_(x->oo) cosx/x`

by applying the squeeze theorem we can solve the limits

and it is as follows,

as we know the boundaries of the `cos(x)` as

`-1<=cos(x)<=1`

now dividing the above expression with `x` , we get

`-1/x<=cos(x)/x<=1/x`

now applying the limits of x-> oo for the above expression, we get

`lim_(x->oo) (-1/x)<=lim_(x->oo) cos(x)/x<=lim_(x->oo) (1/x)`

now upon `x-> oo` we get

`lim_(x->oo) (-1/x) =(-1/oo) =0`

and

`lim_(x->oo) (1/x) = (1/oo)=0`

so,

`lim_(x->oo) (-1/x)<=lim_(x->oo) cos(x)/x<=lim_(x->oo) (1/x)`

=>`0<=lim_(x->oo) cos(x)/x<=0`

=>`lim_(x->oo) cos(x)/x =0`

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