`lim_(z-gtoo)(3z-2)/(z^2-6z+1)^(1/2)`

If we try to apply limits straight away, you will get, oo/oo situation, which is

Now we can divide bothe numerator and denominator by z.

`lim_(z-gtoo)((3z-2)/z)/((z^2-6z+1)^(1/2)/z)`

`lim_(z-gtoo)((3z-2)/z)/((z^2-6z+1)/z^2)^(1/2)`

When we take z into the squareroot value, z is squared.

`lim_(z-gtoo)((3-2/z))/(1-6/z+1/z^2)^(1/2)`

Now if we apply limits as `z= oo` you will get,

`lim_(z-gtoo)((3-2/z))/(1-6/z+1/z^2)^(1/2) = (3 -2/oo)/(1-6/oo+1/(oo)^2)^(1/2)`

This will be reduced to,

`lim_(z-gtoo)((3-2/z))/(1-6/z+1/z^2)^(1/2) = (3-0)/(1-0+0)^(1/2)`

`lim_(z-gtoo)((3-2/z))/(1-6/z+1/z^2)^(1/2) = 3/1 = 3`

Therefore,

`lim_(z-gtoo)(3z-2)/(z^2-6z+1)^(1/2) = 3`