# Calculating limit of a given function

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 !

• Hint: Examine the logarithm of your expression. – uniquesolution Sep 19 '15 at 17:06

$$x^{1 + \ln x} = e^{(1 + \ln x) \ln x}$$
So your expression equals $$e^{\ x^k - (1 + \ln x) \ln x}$$
• @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 – Ant Sep 20 '15 at 11:51