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Got stuck in calculating of a limit $$\lim_{x \to \infty}\frac{\Bbb e^{x^k}}{{x^{1 + \ln x} }}.$$ I was thinking of using Taylor series expansion of exponential, but could not get anything. Then I tried using L' Hospital's which was again useless. I was also thinking solving the limit using some asymptotic function, but could not think of any such function. Guys I need some help !

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    $\begingroup$ Hint: Examine the logarithm of your expression. $\endgroup$ – uniquesolution Sep 19 '15 at 17:06
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Hint

$$x^{1 + \ln x} = e^{(1 + \ln x) \ln x}$$

So your expression equals $$e^{\ x^k - (1 + \ln x) \ln x}$$

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  • $\begingroup$ Now it becomes of the form e^{\infty - \infty} form as x \to \infty. Again an indeterminate form. $\endgroup$ – anthony Sep 20 '15 at 9:27
  • $\begingroup$ Here k is a positive real number. $\endgroup$ – anthony Sep 20 '15 at 9:31
  • $\begingroup$ @anthony yes but the logarithm can be ignored when there are powers of $x$. that is, $x^k - \ln x \sim x^k$. In any case if you want to calculate that limit, write that as $x^k(1 - \ln x(1+\ln x)/x^k)$ and use hopital to the last term inside the parenthesis $\endgroup$ – Ant Sep 20 '15 at 11:51

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