# How $\displaystyle \lim_{x \rightarrow \infty} \frac{x^c}{c^x}=0$, for $c>1$

How $\displaystyle \lim_{x \rightarrow \infty} \frac{x^c}{c^x}=0$, for $c>1$,

L'Hospital's rule doesn't seem to work.

• Some intuition: every time you double the value of $x$, the numerator is multiplied by a factor $2^c$. The denominator is squared. Once the two become big enough, one of those operations is a lot more powerful than the other. – Arthur Nov 17 '14 at 4:36
• Try to use logarithm. – Falang Nov 17 '14 at 4:36
• @Arthur thanks for intuitive insight. – kaka Nov 17 '14 at 4:40
• @Falang yes, as Logarithm(of above function) is approaching to a more negative number, that means that original function is approaching to zero. Thanks. – kaka Nov 17 '14 at 4:45

Hint: $x^c = e^{c\ln x}, c^x = e^{x\ln c}$