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The question is $\lim\limits_{x\rightarrow 0^+} x^{8 \sin(x)}$. It says, use L'Hospital's rule if necessary. Are there other methods to solve this? L'Hospital's rule would be complicated to evaluate, and might lead to other indeterminate forms. I tried to take the natural log and solve, but I failed. Any other methods I could use? I would ike a step-by step walk-through to see if it is possible to use L'Hospital's rule?

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  • $\begingroup$ Sometimes, you have to apply L'H multiple times, checking each time if you have an indeterminate form. $\endgroup$ – Jonathan Dewein Apr 19 '13 at 15:15
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We know the elementary result in analysis $$\sin x\sim_0 x\qquad\mathrm{and}\qquad \lim_{x\to 0^+}x\log x=0$$ then we have $$\lim\limits_{x\rightarrow 0^+}x^{8\sin x}=\lim\limits_{x\rightarrow 0^+}e^{8(\sin x )(\log x)}=\lim\limits_{x\rightarrow 0^+}e^{8x\log x}=e^0=1$$

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$$\lim\limits_{x\rightarrow 0^+}x^{8\sin x}=\lim\limits_{x\rightarrow 0^+}e^{8(\sin x )(\log x)}$$ Now $$\lim\limits_{x\rightarrow 0^+}(\sin x \log x)\geq \lim\limits_{x\rightarrow 0^+}x\log x=\lim\limits_{x\rightarrow 0^+}\dfrac{\log x}{\frac{1}{x}}=(DL'H)$$ $$=\lim\limits_{x\rightarrow 0^+}\dfrac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim\limits_{x\rightarrow 0^+}(-x)=0$$

Also again using De L'Hospital : $$\lim\limits_{x\rightarrow 0^+}(\sin x \log x)\leq \lim\limits_{x\rightarrow 0^+}(\frac{2}{\pi}x\log x)=0$$ Hence: $\lim\limits_{x\rightarrow 0^+}(\sin x \log x)=0 $

Therefore: $$\lim\limits_{x\rightarrow 0^+}x^{8\sin x}=\lim\limits_{x\rightarrow 0^+}e^{8(\sin x )(\log x)}=e^{8\cdot 0}=1$$

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HINT:

$$\text{ Let } y=x^{\sin x}\implies \ln y=\frac{\ln x}{\csc x}$$

Now, $\lim_{x\to0^+}\frac{\ln x}{\csc x}$ is of the form $\frac \infty\infty$

So, applying L'Hospital's rule $$\lim_{x\to0^+} \ln y=\lim_{x\to0^+}\frac{\frac1x}{-\csc x\cot x}=\frac{-1}{\lim_{x\to0^+}\cos x}\cdot\left(\lim_{x\to0^+}\frac{\sin x}x\right)^2\cdot \lim_{x\to0^+}x=0$$ as $\lim_{x\to0}\frac{\sin x}x=1$

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