We houden rekening met de formule   a = e^{ln a}  zodat ook bv. x^x = e^{x.ln x}
Dit levert :
       \(\large\displaystyle\lim_{x\to 0}\:(\cos x)^{\frac{1}{x^2}}\:=\displaystyle\lim_{x\to 0}\:e^{\frac{\ln\cos x}{x^2}}\)
We gaan nu de limiet zoeken van de exponent (grondtal e is constant).
\(\displaystyle\lim_{x\to\,0}\:\frac{\ln\cos x}{x^2}\left(=\frac00\right)\overset{H}{=}\lim_{x\to}\:\frac{\frac{1}{\cos x}(-\sin x)}{2x}=-\frac12\lim_{x\to0}\:\frac{\tan x}{x}=-\frac 12\)
\(\text{Bijgevolg is}\quad\displaystyle\lim_{x\to 0}\:(\cos x)^\frac{1}{x^2}=e^{-\frac12}=\frac{1}{\sqrt e}\;(\approx 0,6)\)