Find the limit: limx^sinx as x approaches to 0+.

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beckden | High School Teacher | (Level 1) Educator

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`lim_(x->0^+) x^sinx` this is a `0^0` type limit,

we take the use the fact that `lim_(x->0^+)f(x)=e^(ln(lim_(x->0^+)f(x)))`

and `ln(lim_(x->0^+)f(x))=lim_(x->0^+)ln(f(x))` to get

`ln(lim_(x->0^+)x^sinx)=lim_(x->0^+)ln(x^sinx))=lim_(x->0^+)sinx*ln(x)`

sinx*lnx is a 0*oo type limit

`lim_(x->0^+)sinx*lnx=lim_(x->0^+)lnx/cscx` Now we can use L'Hopital's rule

to get

`lim_(x->0^+)lnx/cscx=lim_(x->0^+)(1/x)/(-cscx*cotx)`

Now `1/x*1/(-cscx*cotx)=-(sinx*sinx)/(xcosx)`

finally `lim_(x->0)sin^2x/(xcosx) = lim_(x->0)sinx/x * lim_(x->0)(sinx/cosx)=1*0=0`

` ``lim_(x->0^+)x^sinx=``e^(lim_(x->0^+)sinx*lnx)=e^0=1`

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