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

The answer is

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

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To find the limit of ( \sin(x)^{\sin(x)} ) as ( x ) approaches ( 0^+ ), we need to consider the behavior of ( \sin(x)^{\sin(x)} ) as ( x ) approaches ( 0 ) from the right side.

As ( x ) approaches ( 0^+ ), ( \sin(x) ) approaches ( 0 ), and ( \sin(0) = 0 ). Therefore, ( \sin(x)^{\sin(x)} ) approaches ( 0^0 ) as ( x ) approaches ( 0^+ ).

The expression ( 0^0 ) is an indeterminate form, meaning its value is not immediately determinable. However, as ( x ) approaches ( 0^+ ), ( \sin(x) ) approaches ( 0 ), and ( \sin(0) = 0 ). So, ( \sin(x)^{\sin(x)} ) approaches ( 0^0 ).

To resolve the indeterminate form ( 0^0 ), we can consider taking the natural logarithm of the expression and then finding the limit:

[ \lim_{x \to 0^+} \sin(x)^{\sin(x)} = e^{\lim_{x \to 0^+} \sin(x) \ln(\sin(x))} ]

Now, as ( x ) approaches ( 0^+ ), both ( \sin(x) ) and ( \ln(\sin(x)) ) approach ( 0 ). Therefore, ( \lim_{x \to 0^+} \sin(x) \ln(\sin(x)) = 0 ).

So, the original limit becomes:

[ e^0 = 1 ]

Hence, the limit of ( \sin(x)^{\sin(x)} ) as ( x ) approaches ( 0^+ ) is ( 1 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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