# Given that `lim_(x->0)(1-cosx)/x^2=1/2` , quickly evaluate `lim_(x->0)sqrt( 1-cosx)/x`

It is given that `lim_(x->0)(1 - cos x)/x^2=1/2 `

The value of `lim_(x-> 0) sqrt(1 - cos x)/x` has to be determined.

`lim_(x-> 0) sqrt(1 - cos x)/x`

=> `lim_(x-> 0) sqrt((1 - cos x)/x^2)`

=> `sqrt(lim_(x->0)(1 - cos x)/x^2)`

=> `sqrt(1/2)`

=> `1/sqrt 2`

The value of `lim_(x-> 0) sqrt(1 - cos x)/x = 1/sqrt 2`

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You need to remember the half of angle formula such that:

`sqrt(1-cos x) = sqrt2*sin(x/2)`

You should substitute `sqrt2*sin(x/2)`  for `sqrt(1-cos x)`  in limit such that:

`lim_(x-gt0) sqrt2*sin(x/2)/x`

You need to use special limit `lim_(x-gt0) (sin x)/x = 1`  such that:

`lim_(x-gt0) (sqrt2)/2*sin(x/2)/(x/2) = (sqrt2)/2*lim_(x-gt0) sin(x/2)/(x/2)`

`lim_(x-gt0) (sqrt2)/2*sin(x/2)/(x/2) =(sqrt2)/2* 1`

`lim_(x-gt0) (sqrt2)/2*sin(x/2)/(x/2) = (sqrt2)/2`

Hence, evaluating the limit of the function yields `lim_(x-gt0)sqrt(1-cos x)/x = (sqrt2)/2.`

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