We houden rekening met de formule   a = e^{ln a} zodat ook bv. x^x = e^{x.ln x}
en dus ook :
        \(\displaystyle\lim_{x\to 0}\:(\cos x)^{\frac{1}{\sin^{{\small 2}}x}}\:=\displaystyle\lim_{x\to 0}\:e^{\frac{\ln\cos x}{\sin^2x}}\)
We gaan nu de limiet zoeken van de exponent (grondtal e is constant).
\(\displaystyle\lim_{x\to 0}\:\frac{\ln\cos x}{\sin^2x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\frac{\frac{1}{\cos x}(-\sin x)}{2\sin x\cos x}\\=-\displaystyle\lim_{x\to 0}\:\frac{\sin x}{\cos x.2\sin x.\cos x}=-\displaystyle\lim_{x\to 0}\:\frac{1}{2\cos^2x}=-\frac{1}{2.1^2}=-\frac12\)
\(\text{Bijgevolg is}\quad\displaystyle\lim_{x\to 0}\:(\cos x)^{\frac{1}{\sin^{{\small 2}}x}}\:=e^{-\frac12}=\frac{1}{\sqrt e}\;(\approx 0,6)\) ^{gricha - v7485 - 1.10.13}