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

Question

Emilia Aguilar · Accepted Answer

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

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]}

Christopher Allers · Answer

undefined 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 $.