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How can I show the two limits $$ \displaylines{ \mathop {\lim }\limits_{x \to + \infty } \frac{{x^2 e^{x + \frac{1}{x}} }}{{e^{ - x} \left( {\ln x} \right)^2 \sqrt x }} = \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{x^{\frac{3}{4}} e^{\left( {x + \frac{1}{{2x}}} \right)} }}{{\ln x}}} \right)^2 = + \infty \cr \mathop {\lim }\limits_{x \to 0^ + } \left( {\frac{{x^{\frac{3}{4}} e^{\left( {x + \frac{1}{{2x}}} \right)} }}{{\ln x}}} \right)^2 = + \infty \cr} $$ think you

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$$\frac{x^2e^x}{e^{-x} (\ln x)^2\sqrt x} < \frac{x^2e^{x + \frac1x}}{e^{-x} (\ln x)^2\sqrt x} \text{ for large $x$ since } e^{\frac1x} > 1$$

$$$$ $$\lim_{x \to \infty} \frac{x^2e^x}{e^{-x} (\ln x)^2\sqrt x} = e^{2x}\frac{x^{1.5}}{(\ln x)^2} = +\infty$$ $$$$

$$\text{Thus, the right hand side at the top should diverge to $+\infty$ too.}$$

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  • $\begingroup$ and the other limit as x approaches zero $\endgroup$ – sabachir Jan 9 '15 at 20:09
  • $\begingroup$ @sabachir I tried and failed. $\endgroup$ – Mohamad Ali Baydoun Jan 9 '15 at 21:09
  • $\begingroup$ is can I use the limit developments $\endgroup$ – sabachir Jan 10 '15 at 6:19

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