Question

Limit x tends to 0+ `sinx^sinx` ?

Answer 1

The answer is `lim_{x->0^{+}}(sin(x))^{sin(x)}=1`.

Explanation:

First, let `y=(sin(x))^{sin(x)}`. Then `ln(y)=sin(x)ln(sin(x))=(ln(sin(x)))/csc(x)`. Now use L'Hopital's Rule to evaluate the limit of this expression (it is an `infty/infty` indeterminate form).

`lim_{x->0+} (ln(sin(x)))/csc(x)=lim_{x->0+}(1/sin(x) * cos(x))/(-cot(x)*csc(x))`

`=lim_{x->0+}(-tan(x))=0`

Therefore, `ln(lim_{x->0+}y)=lim_{x->0}ln(y)=lim_{x->0+}sin(x)ln(sin(x))=0`. Exponentiation now implies that

`lim_{x->0+}(sin(x))^{sin(x)}=lim_{x->0+}y=e^{0}=1`.

The graph of `(sin(x))^{sin(x)}` confirms this visually.

graph{sin(x)^(sin(x)) [-4.054, 4.065, -2.034, 2.025]}