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Given any $\alpha > 0$, I need to show that for $ x \in [0,\infty)$ \begin{equation} \lim_{x\to 0} x^{\alpha}e^{|\log x|^{1/2}}=0 \end{equation}

I have tried using L'Hospital's rule. But I am not able to arrive at answer.

Thank you in advance.

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  • $\begingroup$ Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $\LaTeX$. $\endgroup$ – hardmath Aug 4 '18 at 15:42
  • $\begingroup$ @hardmath: we can write it as a ratio with exponential term in the denominator. $\endgroup$ – Rahul Raju Pattar Aug 4 '18 at 15:45
  • $\begingroup$ What is $\log x$ for $x<0?$ $\endgroup$ – zhw. Aug 4 '18 at 16:58
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Let $f(x) =x^{a}e^{|\log x|^{1/2}} $. $\ln f(x) =a\ln x+|\log x|^{1/2} $.

Let $x = 1/y$, so $y \to \infty$ as $x \to 0$.

$\ln f(1/y) =a\ln (1/y)+|\log (1/y)|^{1/2} =-a\ln (y)+|\log y|^{1/2} $.

The key is that $\dfrac{|\log y|^{1/2}}{\ln(y)} \to 0$ as $y \to \infty$.

Therefore $\ln f(1/y) =\ln (y)(-a +\dfrac{|\log y|^{1/2}}{\ln(y)}) $. Since $\dfrac{|\log y|^{1/2}}{\ln(y)} \to 0$ as $y \to \infty$ and $a > 0$, $-a +\dfrac{|\log y|^{1/2}}{\ln(y)} \lt -a/2$ for large enough $y$ so that $\ln f(1/y) \to -\infty$ so $f(1/y) \to 0$.

Note that this works for any exponent less than $1$, not just $\frac12$.

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  • $\begingroup$ "Let $x = 1/y$, so $y \to \infty$ as $x \to 0$." Not so, we need $x\to 0^+.$ $\endgroup$ – zhw. Aug 4 '18 at 17:22
  • $\begingroup$ Then $y \to \infty^+$. $\endgroup$ – marty cohen Aug 4 '18 at 22:42
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Let $$f(x) = x^{\alpha} e^{\sqrt{\vert \log x \vert}}$$ Consider $$\log f(x) = \alpha \log x + \sqrt{\vert \log x \vert} = \log x (\alpha + \frac{ \sqrt{\vert \log x \vert}}{\log x} )=\log x (\alpha - \frac{ \sqrt{\vert \log x \vert}}{\sqrt{\vert \log x \vert}\sqrt{\vert \log x \vert}} )$$ which is $$\log f(x) = \log x (\alpha - \frac{ 1}{\sqrt{\vert \log x \vert}} )$$ As $x$ goes to zero $\sqrt{\vert \log x \vert}$ goes to $+ \infty$ hence $\alpha - \frac{ 1}{\sqrt{\vert \log x \vert}}$ goes to $\alpha$. Hence

$$\lim_{x \rightarrow 0}\log f(x) =\lim_{x \rightarrow 0} \alpha \log x = -\infty$$ So $$\lim_{x \rightarrow 0} f(x) = e^{-\infty} = 0$$

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Let $x=e^{-y}\to 0$ as $y\to \infty$ then

$$\large{x^{\alpha}e^{|\log x|^{1/2}}=e^{-\alpha y}\,e^{\sqrt y}=e^{\sqrt y\,-\,\alpha y}\to 0}$$

indeed

$$\sqrt y-\alpha y=y\left(\frac {\sqrt y}y -\alpha\right)=y\left(\frac 1 {\sqrt y}-\alpha\right) \to -\infty$$

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