# Solve limit using L'Hopital's rule: lim(2/pi arccos x)^(1/x), if x->0

We can use that `lim_(x->0) f(x) = lim_(x->0) e^(ln(f(x))) = e^(lim_(x->0+)ln(f(x)))`

`lim_(x->0) (2/pi arccos(x))^(1/x) = lim_(x->0) e^(ln((2/pi arccos(x))^(1/x)))`

`lim_(x->0)ln(2/pi arccos(x))^(1/x)=lim_(x->0)ln(2/pi arccos(x))/x`

Now we use L'Hopital's rule because

`lim_(x->0) ln(2/pi arccos(x)) = 0` and `lim_(x->0) x = 0` so we have `0/0` and can use L'Hopital

`lim_(x->0)ln(2/pi arccos(x))/x=lim_(x->0)(1/(2/pi arccos(x))(2/pi (-1/sqrt(1-x^2))))/1`

This...

We can use that `lim_(x->0) f(x) = lim_(x->0) e^(ln(f(x))) = e^(lim_(x->0+)ln(f(x)))`

`lim_(x->0) (2/pi arccos(x))^(1/x) = lim_(x->0) e^(ln((2/pi arccos(x))^(1/x)))`

`lim_(x->0)ln(2/pi arccos(x))^(1/x)=lim_(x->0)ln(2/pi arccos(x))/x`

Now we use L'Hopital's rule because

`lim_(x->0) ln(2/pi arccos(x)) = 0` and `lim_(x->0) x = 0` so we have `0/0` and can use L'Hopital

`lim_(x->0)ln(2/pi arccos(x))/x=lim_(x->0)(1/(2/pi arccos(x))(2/pi (-1/sqrt(1-x^2))))/1`

This limit can be evaluated at x=0

`1/(2/pi arccos(0))(2/pi (-1/sqrt(1-0)))=-2/pi`

So our `lim_(x-gt0) (2/pi arccos(x))^(1/x)=e^(-2/pi)`

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