Question

How will I solve this limit? lim n----> infinity n sin(π/4n)

Answer 1

`lim_(n->oo)nsin(pi/(4n))=pi/4`

Explanation:

We know limit of `lim_(x->0)sinx/x=1`

Here we have `lim_(n->oo)nsin(pi/(4n))`, which can also be written as

`lim_(n->oo)sin(pi/(4n))/(1/n)`

or `lim_(n->oo)sin(pi/(4n))/(1/n*pi/4)*pi/4`

or `lim_(n->oo)sin(pi/(4n))/(pi/(4n))*pi/4`

Now if we let `(pi/(4n))=x`, as `n->oo`, `x->0`

and our limit is `lim_(x->0)sinx/x*pi/4=pi/4`