I tried writing $y=x^{\sin x}$, so $\ln y=\sin x\ln x$. I tried to rewrite the function as $\ln(x)/\csc x$ and apply l'Hopital to the last function, but it's a mess.
Is there a by hand way to do it?
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Your first approach using $\ln$ is good.
So set $$\ln y = \sin x \ln x$$
As $x\to 0^+$, $\sin x \approx x$ therefore: $$\ln y \approx x\ln x \to 0$$ as $x\to 0^+$.
So $y\to 1$
user88595
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Your idea is right. May be you made a mistake in your calculations. Here is what I have $$ \lim_{x\to0} \frac{\ln x}{1/\sin x} = \lim_{x\to0} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} (1/\sin x)} = \lim_{x\to0}\frac{1/x}{-\cos x/\sin^2x} =\lim_{x\to0} -\frac{\sin^2x}{x\cos x}$$ The last limit can be written as $$ \lim_{x\to0} -\frac{\sin^2x}{x\cos x} =\lim_{x\to0} -x \left[\frac{\sin x}{x}\right]^2\frac 1{\cos x} = 0 \cdot 1^2 \cdot 1 = 0 $$ So $$\ln y \to 0 \Rightarrow y\to 1$$
user44197
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But $\lim_{x \rightarrow 0} \frac{1}{\sin x}$ do not exist and you cannot apply l'Hopital – Matheman Jan 06 '14 at 21:17
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the ratio is $\infty/\infty$ and you can use L'hopital rule in this case also. – user44197 Jan 06 '14 at 21:27
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I'm not sure. See the L'hopital theorem statement – Matheman Jan 06 '14 at 21:33
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I have been using L'hopital rule for both $0/0$ and $\infty/\infty$. It works for both cases. – user44197 Jan 06 '14 at 21:42
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Ok,sorry! You're right! – Matheman Jan 06 '14 at 21:43
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$\sin x \sim x,$ near $0,$ so you want to find the limit of $x \log x.$ Write it (as you did) as $\log x/(1/x),$ and use L'hopital.
Igor Rivin
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Or also
$$\lim_{x\rightarrow 0} x^{\frac{\sin x}{x}x}$$
since $\lim_{x\rightarrow 0} \frac{\sin x}{x}=1$ and ${0}^0=1$
Matheman
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3This is wrong, since $(x,y)\mapsto x^y$ is not continuous at $(0,0)$. – Carsten S Jan 06 '14 at 20:58
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3If you simply use the fact that $0^0 = 1$, why do you go through $\frac{\sin x}{x}x$? – user88595 Jan 06 '14 at 21:00
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@ user88595 $\frac{\sin x}{x}x \rightarrow 1 \cdot 0$ as $x \rightarrow 0$ – Matheman Jan 06 '14 at 21:24
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If you claim that $a_n^{b_n}\to1$ whenever $a_n\to0$ and $b_n\to0$, then it is wrong. If you claim that $a_n^{b_n}\to1$ whenever $a_n\to 0$ and $b_n/a_n\to1$, then you still have to prove it. – Carsten S Jan 06 '14 at 22:22
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@twinprime : Yes but $\sin x\to 0$ as $x\to 0$. I don't get the divided and multiplied by $x$... – user88595 Jan 07 '14 at 12:48
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Write $x^{\sin{x}} = \exp{\frac{\log{x}}{\frac{1}{\sin{x}}}}$ and apply rule of l'hopital:
$$\lim_{x\rightarrow 0}x^{\sin{x}} = \exp{\left(\lim_{x\rightarrow 0}\frac{\log{x}}{\frac{1}{\sin{x}}}\right)}$$
Now (by 2$\times$ l'hopital)$$\lim_{x\rightarrow 0}\frac{\log{x}}{\frac{1}{\sin{x}}} = \lim_{x\rightarrow 0}\frac{\sin{2x}}{ \cos{x}-x\sin{x}} = 0$$
Thus $$\lim_{x\rightarrow 0}x^{\sin{x}} = \exp{\left(\lim_{x\rightarrow 0}\frac{\sin{2x}}{ \cos{x}-x\sin{x}}\right)} = \exp{0} = 1$$
user127.0.0.1
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Rewrite the limit of the logarithm as $$\lim_{x\to 0^+}\frac{\sin x}{x}x\ln x.$$ The behaviour of $\frac{\sin x}{x}$ is known. So we need $\lim_{x\to 0^+} x\ln x$, which is probably familiar, and yields to one application of L'Hospital's Rule.
André Nicolas
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Another way to think about this:
$$\lim_{x \to 0} \sin x \ln x = \lim_{x \to 0} x \ln x$$ (because $(\sin x) /x \to 1$ as $x \to 0$). Writing $x = 1/y$, we have $$\lim_{x \to 0} x \ln x = - \lim_{y \to +\infty}(\ln y)/y.$$ Now $\ln y$ grows slower than any positive power of $y$, and so $\ln y / y \to 0$ as $y \to +\infty$.
(Or, one can make another subsitution: $y = e^z,$ so that $(\ln y)/y = z/e^z,$ and then use the fact that $e^z$ grows faster than any power of $z$, so that $z/e^z \to 0$ as $z \to + \infty$.)
Matt E
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