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

Given to solve,

`lim_(x->oo) sinx/(x-pi)`

This can be solved by applying the squeeze theorem and is as follows

as we know the limits or boundaries of `sin(x)` is

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

Dividing the above expression with x-pi we get

`-1/(x-pi)<=sin(x)/(x-pi)<=1/(x-pi)`

now, let us apply the limits that` x-> oo` we get

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

but

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

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

so,

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

so ,

`lim_(x->oo) sin(x)/(x-pi) = 0`

Given to solve,

`lim_(x->oo) sinx/(x-pi)`

This can be solved by applying the squeeze theorem and is as follows

as we know the limits or boundaries of `sin(x)` is

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

Dividing the above expression with x-pi we get

`-1/(x-pi)<=sin(x)/(x-pi)<=1/(x-pi)`

now, let us apply the limits that` x-> oo` we get

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

but

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

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

so,

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

so ,

`lim_(x->oo) sin(x)/(x-pi) = 0`