Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: (\cos x)^{\frac {1} {\sin^{{\small 2}} x}}\:=$	A. $1$
B. $e$
C. $\frac{1}{e}$
D. $\sqrt e$
E. $\frac{1}{\sqrt e}$

$https://latex.codecogs.com/png.image?\dpi{110}\displaystyle \lim_{x \to 0} \: (\cos x)^{\frac {1} {\sin^{{\small 2}} x}}\: =$	A. $1$
	B. $e$
	C. $\frac{1}{e}$
	D. $\sqrt e$
	E. $\frac{1}{\sqrt e}$

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).
$\displaystyle\lim_{x\to 0}\:\frac{\ln\cos x}{\sin^2x}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 0}\:\frac{\frac{1}{\cos x}(-\sin x)}{2\sin x\cos x}\\=-\displaystyle\lim_{x\to 0}\:\frac{\sin x}{\cos x.2\sin x.\cos x}=-\displaystyle\lim_{x\to 0}\:\frac{1}{2\cos^2x}=-\frac{1}{2.1^2}=-\frac12$
$\text{Bijgevolg is}\quad\displaystyle\lim_{x\to 0}\:(\cos x)^{\frac{1}{\sin^{{\small 2}}x}}\:=e^{-\frac12}=\frac{1}{\sqrt e}\;(\approx 0,6)$