Question

How do you find `lim_(x->0^+)(sinx)^x`?

How do you find `lim_(x->0^+)(sinx)^x`?

Answer 1

`lim_(xrarr0^+)sinx^x=1`

Explanation:

`lim_(xrarr0^+)sinx^x=?`

`sinx^x=e^(ln(sinx)^x)=e^(xln(sinx))`

`= lim_(xrarr0^+)e^(xln(sinx))` `(1)`

You see that we still have problem and can't calculate the limit at it's current form as `ln0` is not defined.
Let's study this limit seperately

`lim_(xrarr0^+)ln(sinx)=`

Set

`sinx=u`
`x->0^+`
`u->0^+`

`=` `lim_(urarr0^+)lnu=-oo`

graph{lnx [-10, 10, -5, 5]}

`lim_(xrarr0^+)xln(sinx)=lim_(xrarr0^+)ln(sinx)/(1/x)=_(DLH)^(((-oo)/(+oo)))`

`lim_(xrarr0^+)(((sinx)')/sinx)/(-1/x^2)=-lim_(xrarr0^+)(cosx/sinx)/(1/x^2)=-lim_(xrarr0^+)((x^2cosx)/sinx)`

`x->0^+`

`x>0`

`=` `-lim_(xrarr0^+)(cosx)/(sinx/x*1/x)=^((1/(1*(+oo)))` `0`

• because

`lim_(xrarr0^+)cosx=cos0=1`

`lim_(xrarr0^+)sinx/x=1`

`lim_(xrarr0^+)(1/x)=^((1/(0^+))` `+oo`

Now if in `(1)` i set:

`y=xln(sinx)`

`x->0^+`
`y->0`

and `(1)` now becomes:

`lim_(yrarr0)e^y=e^0=1`

As a result,

`lim_(xrarr0^+)sinx^x=1`