L=\lim_(x \to \0) x^(sin x)

Question

L=
`\lim_(x \to \0) x^(sin x)`

Answer 1

By applying the exponential and logarithmic functions, we have

`x^(\sin(x)) = \exp \left(\ln \left(x^(\sin(x))\right)\right)`

Then in the limit,

`\displaystyle \lim_(x\to0) x^(\sin(x)) = \lim_(x\to0) \exp \left(\ln \left(x^(\sin(x))\right)\right)`

`\displaystyle \lim_(x\to0) x^(\sin(x)) = \exp \left( \lim_(x\to0) \ln \left(x^(\sin(x))\right)\right)`

`\displaystyle \lim_(x\to0) x^(\sin(x)) = \exp \left( \lim_(x\to0) \sin(x) \ln(x) \right)`

`\displaystyle \lim_(x\to0) x^(\sin(x)) = \exp \left( \lim_(x\to0) (\ln(x))/(\csc(x)) \right)`

As x approaches 0 (from the right), both ln(x) and csc(x) approach infinity (ignoring sign). Applying L'Hopitâl's rule gives

`\displaystyle \lim_(x\to0) x^(\sin(x)) = \exp \left( \lim_(x\to0) (\frac1x)/(-\csc(x)\cot(x)) \right) = \exp \left( -\lim_(x\to0) (\sin^2(x))/(x\cos(x)) \right)`

Recall that

`\displaystyle \lim_(x\to0) (\sin(x))/(x) = 1`

Then

`\displaystyle \lim_(x\to0) (\sin^2(x))/(x\cos(x)) = \lim_(x\to0) (\sin(x))/(\cos(x)) = \lim_(x\to0) \tan(x) = 0`

So, our limit is

`\displaystyle \lim_(x\to0) x^(\sin(x)) = \exp(0) = \boxed{1}`