80.4k views
5 votes
L=
\lim_(x \to \0) x^(sin x)

1 Answer

5 votes

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}

User Sam Warwick
by
8.8k points

Related questions

asked Feb 2, 2024 161k views
Linusz asked Feb 2, 2024
by Linusz
7.9k points
1 answer
2 votes
161k views
1 answer
4 votes
5.7k views
1 answer
1 vote
133k views