How do i know what is the limit :
$$ \lim_{x \to \infty} \frac {e^{x/2}}{2^x} $$
Because with l'hopital it is imposible to say, as the nth derivatives of both $ {e^{x/2}} $and $ {2^x}$ are always infinity...
I could use a hint that helps me think this through
Answer
$$\large\frac{e^{x/2}}{2^x}=\frac{e^{x/2}}{e^{x\ln(2)}}=e^{x\left[\frac12-\ln(2)\right]}\to\ ...$$
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