find limit 3^sinx-1/x as x approaches 0? please help

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if your problem is `lim_(x->0)(3^(sin(x))-1)/x`

We can use L'Hopital's rule because `3^(sin(x))-1 -> 0` and `x->0` as `x->0`

so we can use `lim_(x->0)(f(x))/(g(x))=lim_(x->0)(f'(x))/(g'(x))`

if `f(x)=3^(sin(x))-1` then `f'(x)=ln(3)3^(sin(x))cos(x)=ln(3)cos(x)3^(sin(x))`

and if `g(x)=x` so `g'(x)=1` so

`lim_(x->0)(3^(sin(x))-1)/x=lim_(x->0)(ln(3)cos(x)3^(sin(x)))/1=ln(3)cos(0)3^(sin(0))```

so our answer is

`lim_(x->0)(3^(sin(x))-1)/x=ln(3)`

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if your problem is `lim_(x->0)(3^(sin(x))-1)/x`

We can use L'Hopital's rule because `3^(sin(x))-1 -> 0` and `x->0` as `x->0`

so we can use `lim_(x->0)(f(x))/(g(x))=lim_(x->0)(f'(x))/(g'(x))`

if `f(x)=3^(sin(x))-1` then `f'(x)=ln(3)3^(sin(x))cos(x)=ln(3)cos(x)3^(sin(x))`

and if `g(x)=x` so `g'(x)=1` so

`lim_(x->0)(3^(sin(x))-1)/x=lim_(x->0)(ln(3)cos(x)3^(sin(x)))/1=ln(3)cos(0)3^(sin(0))```

so our answer is

`lim_(x->0)(3^(sin(x))-1)/x=ln(3)`

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