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My question is, what is $\lim_{x \rightarrow 0^+}(\sin x)^{\ln x}$?

This limit is equal to

$$\lim_{x \rightarrow 0^+} e^{\ln(\sin x)\ln x} = e^{\lim_{x\to0^+} \ln(\sin x)\ln x}$$

But what is right hand side limit of $\ln(\sin x)\ln x$ when x goes to $0$

any suggestion? I cannot apply L'Hospital.

Clement C.
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corcia candy
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2 Answers2

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Outline:

First, when $x\to 0^+$, we have $\sin x > 0$ for $x$ sufficiently small and the quantity is well-defined. Now (the following will rely on Taylor approximations): $$ \sin x = x + o(x^2) $$ so $$ \ln \sin x = \ln( x + o(x^2)) = \ln x + \ln (1+o(x)) = \ln x + o(x) $$ and $$ e^{\ln( \sin x) \cdot \ln x} = e^{(\ln x)^2 + o(x\ln x)}= e^{(\ln x)^2 + o(1)} $$ as $x\ln x \xrightarrow[x\to 0]{} 0$. To conclude, observing that $(\ln x)^2 \xrightarrow[x\to 0]{} \infty$ will give you the answer by continuity of $\exp$.

Clement C.
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Using a well-known inequality, for $0 < x < 1$ we have

$$1-x < -\ln x < \frac{1- x}{x}. $$

Hence, for $0 < x < 1$

$$\ln x \ln \sin x= (-\ln x )(-\ln \sin x)> -\ln x (1-\sin x)> - \ln x (1 - \sin 1)\to \infty$$

RRL
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  • The upper bound is just concavity. How does the lower bound work? – Ian Nov 18 '15 at 19:05
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    @Ian Those well-known bounds for the logarithm can be obtained in a variety of ways. One of those is to use the integral definition as $$\log x=\int_1^x \frac{1}{u},du$$from which it is clear that $$\frac{x-1}{x}\le \log x \le x-1$$ – Mark Viola Nov 18 '15 at 19:06
  • @Dr. MV: Thats the argument - thanks. I just had the inequality reversed for $\ln x < 0$. Corrected. – RRL Nov 18 '15 at 19:14
  • @RRL Well done. ... +1 – Mark Viola Nov 18 '15 at 19:40